Section 1.1

11. If y=y1=x−2, then y′(x)=−2x−3 and y″(x)=6x−4, so x2y″+5xy′+4y=x2(6x−4)+5x(−2x−3)+4(x−2)=6x−2−10x−2+4x−2=0. If y=y2=x−2ln x, then y′(x)=x−3−2x−3ln x and y″(x)=−5x−4+6x−4ln x, so x2y″+5xy′+4y=x2(−5x−4+6x−4ln x)+5x(x−3−2x−3ln x)+4(x−2ln x)=0.
13. r=23
14. r=±12
15. r=−2, 1
16. r=16(−3±57)
17. C=2
18. C=3
19. C=6
20. C=11
21. C=7
22. C=1
23. C=−56
24. C=17
25. C=π/4
26. C=−π
27. y′=x+y
28. y′=2y/x
29. y′=x/(1−y)
31. y′=(y−x)/(y+x)
32. dP/dt=kP
33. dv/dt=kv2
35. dN/dt=k(P−N)
37. y≡1 or y=x
39. y=x2
41. y=12ex
42. y=cosx or y=sinx
43. (b) The identically zero function x(0)≡0
44. (a) The graphs (figure below) of typical solutions with k=12 suggest that (for each) the value x(t) increases without bound as t increases.

(b) The graphs (figure below) of typical solutions with k=−12 suggest that now the value x(t) approaches 0 as t increases without bound.
45. P(t)=100/(50−t); P=100 when t=49, and P=1000 when t=49.9. Thus it appears that P(t) grows without bound as t approaches 50.
46. v(t)=50/(5+2t); v=1 when t=22.5, and v=110 when t=247.5. Thus it appears that v(t) approaches 0 as t increases without bound.
47. (a) C=10.1; (b) No such C, but the constant function y(x)≡0 satisfies the conditions y′=y2 and y(0)=0.

Section 1.2

1. y(x)=x2+x+3
2. y(x)=13(x−2)3+1
3. y(x)=13(2x3/2−16)
4. y(x)=−1/x+6
5. y(x)=2x+2−5
6. y(x)=13[(x2+9)3/2−125]
7. y(x)=10 tan−1 x
8. y(x)=12 sin 2x+1
9. y(x)=sin−1 x
10. y(x)=−(x+1)e−x+2
11. x(t)=25t2+10t+20
12. x(t)=−10t2−15t+5
13. x(t)=12t3+5t
14. x(t)=13t3+12t2−7t+4
15. x(t)=13(t+3)4−37t−26
16. x(t)=43(t+4)3/2−5t−293
17. x(t)=12[(t+1)−1+t−1]
19. x(t)={5tif 0≤t≤5,10t−12t2−252if 5≤t≤10.
20. x(t)={12t2if 0≤t≤5,5t−252if 5≤t≤10.
21. x(t)={12t2if 0≤t≤5,10t−12t2−25if 5≤t≤10.
22. x(t)={56t2if 0≤t≤3,5t−152if 3≤t≤7,16(−5t2+100t−290)if 7≤t≤10.
23. v(t)=−(9.8)t+49, so the ball reaches its maximum height (v=0) after t=5 seconds. Its maximum height then is y(5)=122.5(m).
24. v(5)=−160 ft/s
25. The car stops when t≈2.78 (s), so the distance traveled before stopping is approximately x(2.78)≈38.58 (m).
26. (a) y≈530 m (b) t≈20.41 s (c) t≈20.61 s
27. y0≈178.57 (m)
28. v(4.77)≈−192.64 ft/s
29. After 10 seconds the car has traveled 200 ft and is traveling at 70 ft/s.
30. a=22 ft/s2; it skids for 4 seconds.
31. v0=1030 (m/s), about 197.18 km/h
32. 60 m
33. 2010≈63.25 (ft/s)
34. 460.8 ft
36. About 13.6 ft
37. 25 (mi)
38. 1:10 pm
39. 6 mph
40. 2.4 mi
41. 5443≈181.33 (ft/s)
42. 25 mi
43. Time: 6.12245×109 s≈194 years;

Distance: 1.8367×1017 m≈19.4 light-years
44. About 54 mi/h

Section 1.3

1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11. A unique solution exists in some neighborhood of x=1.
12. A unique solution exists in some neighborhood of x=1.
13. A unique solution exists in some neighborhood of x=0.
14. Existence but not uniqueness is guaranteed in some neighborhood of x=0.
15. Neither existence nor uniqueness is guaranteed in any neighborhood of x=2.
16. A unique solution exists in some neighborhood of x=2.
17. A unique solution exists in some neighborhood of x=0.
18. Neither existence nor uniqueness is guaranteed.
19. A unique solution exists in some neighborhood of x=0.
20. A unique solution exists in some neighborhood of x=0.
21. Your figure should suggest that y(−4)≈3; an exact solution of the differential equation gives y(−4)=3+e−4≈3.0183.
22. y(−4)≈−3
23. Your figure should suggest that y(2)≈1; the actual value is closer to 1.004.
24. y(2)≈1.5
25. Your figure should suggest that the limiting velocity is about 20 ft/sec (quite survivable) and that the time required to reach 19 ft/sec is a little less than 2 seconds. An exact solution gives v(t)=19 when t=58ln 20≈1.8723.
26. A figure suggests that there are 40 deer after about 60 months; a more accurate value is t≈61.61. The limiting population is 75 deer.
27. The initial value problem y′=2y, y(0)=b has no solution if b<0; a unique solution if b>0; infinitely many solutions if b=0.
28. The initial value problem xy′=y, y(a)=b has a unique solution if a≠0; infinitely many solutions if a=b=0; no solution if a=0 but b≠0.
29. The initial value problem y′=3y2/3, y(a)=b always has infinitely many solutions defined for all x. However, if b≠0 then it has a unique solution near x=a.
30. The initial value problem y′=−1−y2, y(a)=b has a unique solution if |b|<1; no solution if |b|>1, and infinitely many solutions (defined for all x) if b=±1.
31. The initial value problem y′=1−y2, y(a)=b has a unique solution if |b|<1; no solution if |b|>1, and infinitely many solutions (defined for all x) if b=±1.
32. The initial value problem y′=4xy, y(a)=b has infinitely many solutions (defined for all x) if b≧0; no solutions if b<0. However, if b>0 then it has a unique solution near x=a.
33. The initial value problem x2y′+y2=0, y(a)=b has a unique solution with initial point (a, b) if a≠0, no solution if a=0 but b≠0, infinitely many solutions if a=b=0.
34. (a) If y(−1)=−1.2 then y(1)≈−0.48. If y(−1)=−0.8 then y(1)≈2.48. (b) If y(−3)=−3.01 then y(3)≈−1.0343.

If y(−3)=−2.99 then y(3)≈7.0343.
35. (a) If y(−3)=−0.2 then y(2)≈2.019. If y(−3)=+0.2 then y(2)≈2.022. In either case, y(2)≈2.02. (b) If y(−3)≈−0.5 then y(2)≈2.017. If y(−3)≈+0.5 then y(2)≈2.024. In either case, y(2)≈2.02.

Section 1.4

1. y(x)=C exp(−x2)
2. y(x)=1/(x2+C)
3. y(x)=C exp(−cos x)
4. y(x)=C(1+x)4
5. y(x)=sin (C+x)
6. y(x)=(x3/2+C)2
7. y(x)=(2x4/3+C)3/2
8. y(x)=sin−1(x2+C)
9. y(x)=C(1+x)/(1−x)
10. y(x)=(1+x)/[1+C(1+x)]−1
11. y(x)=(C−x2)−1/2
12. y2+1=Cex2
13. ln(y4+1)=C+4 sinx
14. 3y+2y3/2=3x+2x3/2+C
15. 1/(3y3)−2/y=1/x+ln|x|+C
16. y(x)=sec−1(C1+x2)
17. ln|1+y|=x+12x2+C
18. y(x)=tan(C−1x−x)
19. y(x)=2 exp(ex)
20. y(x)=tan(x3+π/4)
21. y2=1+x2−16
22. y(x)=−3 exp(x4−x)
23. y(x)=12(1+e2x−2)
24. y(x)=π2sin x
25. y(x)=x exp(x2−1)
26. y(x)=1/(1−x2−x3)
27. y=ln(3e2x−2)
28. y(x)=tan−1(x−1)
29. (a) General solution y(x)=−1/(x−C); (b) The singular solution y(x)≡0. (c) In the following figure we see that there is a unique solution through every point of the xy-plane.
30. General solution y(x)=(x−C)2; singular solution y(x)≡0. (a) No solution if b<0; (b) Infinitely many solutions (for all x) if b≧0; (c) Two solutions near (a, b) if b>0.
31. Separation of variables gives the same general solution y=(x−C)2 as in Problem 30 , but the restriction that y′=2y≧0 implies that only the right halves of the parabolas qualify as solution curves. In the figure below we see that through the point (a, b) there passes (a) No solution curve if b<0, (b) a unique solution curve if b>0, (c) Infinitely many solution curves if b=0.
32. General solution y(x)=±sec(x−C); singular solutions y(x)≡±1.

(a) No solution if |b|<1; (b) A unique solution if |b|>1; (c) Infinitely many solutions if b=±1.
33. About 51840 persons
34. t ≈3.87 hr
35. About 14735 years
36. Age about 686 years
37. $21103:48
38. $44.52
39. 2585 mg
40. About 35 years
41. About 4.86×109 years ago
42. About 1.25 billion years
43. After a total of about 63 min have elapsed
44. About 2.41 minutes
45. (a) 0.495 m; (b (8.32×10−7)I0; (c) 3.29 m
46. (a) About 9.60 inches; (b) About 18,200 ft
47. After about 46 days
48. About 6 billion years
49. After about 66 min 40 s
50. (a) A(t)=10·32t/15; (b) About 20.80 pu; (c) About 15.72 years
51. (a) A(t)=15·(23)t/5; (b) approximately 7.84 su; (c) After about 33.4 months
52. About 120 thousand years ago
53. About 74 thousand years ago
54. 3 hours
55. 972 s
56. At time t=2048/1562≈1.31 (in hours)
58. 1:20 p.m.
59. (a) y(t)=(8−7t)2/3; (b) at 1:08:34 p.m.; (c) r=160712≈0.15 (in.)
60. About 6 min 3 sec
61. Approximately 14 min 29 s
62. The tank is empty about 14 seconds after 2:00 p.m.
63. (a) 1:53:34 p.m.; (b) r≈0.04442 ft ≈0.53 in.
64. r=17203 ft, about 135 in.
65. At approximately 10:29 a.m.

Section 1.5

1. y(x)=2(1−e−x)
2. y(x)=(3x+C)e2x
3. y(x)=e−3x(x2+C)
4. y(x)=(x+C)ex2
5. y(x)=x+4x−2
6. y(x)=x2+32/x5
7. y(x)=5x1/2+Cx−1/2
8. y(x)=3x+Cx−1/3
9. y(x)=x(7+ln x)
10. y(x)=3x3+Cx3/2
11. y(x)≡0
12. y(x)=14x5−56x−3
13. y(x)=(ex+e−x)/2
14. y(x)=x3 ln x+10x3
15. y(x)=[1−5 exp(−x2)]/2
16. y(x)=1+e−sinx
17. y(x)=(1+sinx)/(1+x)
18. y(x)=x2(sinx+C)
19. y(x)=12sinx+Ccscx
20. y(x)=−1+exp(x+12x2)
21. y(x)=x3 sinx
22. y(x)=(x3+5)ex2
23. y(x)=x3(2+Ce−2x)
24. y(x)=13[1+16(x2+4)−3/2]
25. y(x)=[exp(−32x2)][3(x2+1)3/2−2]
26. x(y)=1/2y2+C/y4
27. x(y)=ey(C+12y2)
28. x(y)=12[y+(1+y2)(tan−1y+C)]
30. y(x)=x1/2∫1xt−1/2cost dt
29. y(x)=[exp(x2)][C+12π erf(x)]
32. (a) y(x)=sinx−cosx; (b) y(x)=Ce−x+sinx−cosx; (c) y(x)=2e−x+sinx−cosx
33. After about 7 min 41 s
34. About 22.2 days
35. About 5.5452 years
36. (a) x(t)=(60−t)−(60−t)3/3600; (b) About 23.09 lb
37. 393:75 lb
38. (a) x(t)=50e−t/20; (b) y(t)=150e−t/40−100e−t/20; (c) 56.25 lb
39. (b) ymax=100e−1≈36.79 (gal)
41. (b) Approximately $1,308,283
43. −50.0529, −28.0265, −6.0000, 16.0265, 38.0529
44. 3.99982, 4.00005, 4.00027, 4.00050, 4.00073
45. x(t)=20(1−e−t/10); x=10 after t=10 ln 2≈6.93 months.
46. x(t)=20101(101−102e−t/10+cos t+10 sin t); x=10 after t=6.47 months.

Section 1.6

1. x2−2xy−y2=C
2. y2=x2(ln x+C)
3. y(x)=x(C+ln |x|)2
4. 2 tan−1(y/x)−ln(y2/x2+1)=2 ln x+C
5. ln |xy|=C+xy−1
6. 2y ln y=x+Cy
7. y3=3x3(C+ln|x|)
8. y=−x ln(C−ln x)
9. y(x)=x/(C− ln|x|)
10. x2+xy2=Cx6
11. y=C(x2+y2)
12. 4x2+y2=x2(ln x+C)2
13. y+x2+y2=Cx2
14. x−x2+y2=C
15. x2(2xy+y2)=C
16. x=2x+y+1−2 ln(1+x+y+1)+C
17. y(x)=−4x+2 tan(2x+C)
18. y=ln(x+y+1)+C
19. y2=x/(2+Cx5)
20. y3=3+Ce−3x2
21. y2=1/(Ce−2x−1)
22. y3=7x/(7Cx7+15)
23. y(x)=(x+Cx2)−3
24. y2=e2x/(C+ln x)
25. 2x3y3=31+x4+C
26. y3=e−x(x+C)
27. y(x)=(x4+Cx)1/3
28. y=ln(Cx2+x2e2x)
29. sin2y=4x2+Cx
30. x2−2xey−e2y=C
31. x2+3xy+y2=C
32. 2x2−xy+3y2=C
33. x3+2xy2+2y3=C
34. x3+x2y2+y4=C
35. 3x4+4y3+12y ln x=C
36. x+exy+y2=C
37. sin x+x ln y+ey=C
38. x2+2x tan−1y+ln(1+y2)=C
39. 5x3y3+5xy4+y5=C
40. exsin y+xtan y=C
41. x2y−1+y2x−3+2y1/2=C
42. xy−2/3+x−3/2y=C
43. y(x)=Ax2+B
44. x(y)=Ay2+B
45. y(x)=A cos 2x+B sin 2x
46. y(x)=x2+A ln x+B
47. y(x)=A−ln |x+B|
48. y(x)=ln x+Ax−2+B
49. y(x)=±(A+Bex)1/2
50. y(x)=ln|sec(x+A)|−12x2+B
51. x(y)=−13(y3+Ay+B)
52. Ay2−(Ax+B)2=1
53. y(x)=A tan(Ax+B)
54. Ay2(B−x)=1
58. y=exp(x2+C/x2)
59. x2−2xy−y2−2x−6y=C
60. (x+3y+3)5=C(y−x−5)
61. x=tan(x−y)+sec(x−y)+C
64. y(x)=x+e−x2[C+12π erf(x)]−1
65. y(x)=x+(C−x)−1
69. Approximately 3.68 mi

Chapter 1 Review Problems

1. Linear: y(x)=x3(C+ln x)
2. Separable: y(x)=x/(3−Cx−xln x)
3. Homogeneous: y(x)=x/(C−ln x)
4. Exact: x2y3+ex−cos y=C
5. Separable: y(x)=Cexp(x−3−x−2)
6. Separable: y(x)=x/(1+Cx+2xln x)
7. Linear: y(x)=x−2(C+ln x)
8. Homogeneous: y(x)=3Cx/(C−x3)=3x/(1+Kx3)
9. Bernoulli: y(x)=(x2+Cx−1)2
10. Separable: y(x)=tan(C+x+13x3)
11. Homogeneous: y(x)=x/(C−3 ln x)
12. Exact: 3x2y3+2xy4=C
13. Separable: y(x)=1/(C+2x2−x5)
14. Homogeneous: y2=x2/(C+2 ln x)
15. Linear: y(x)=(x3+C)e−3x
16. Substitution: v=y−x; solution: y−x−1=Ce2x(y−x+1)
17. Exact: ex+ey+exy=C
18. Homogeneous: y2=Cx2(x2−y2)
19. Separable: y(x)=x2/(x5+Cx2+1)
20. Linear: y(x)=2x−3/2+Cx−3
21. Linear: y(x)=[C+ln(x−1)]/(x+1)
22. Bernoulli: y(x)=(2x4+Cx2)3
23. Exact: xey+y sin x=C
24. Separable: y(x)=x1/2/(6x2+Cx1/2+2)
25. Linear: y(x)=(x+1)−2(x3+3x2+3x+C)=x+1+K(x+1)−2
26. Exact: 3x3/2y4/3−5x6/5y3/2=C
27. Bernoulli: y(x)=x−1(C+ln x)−1/3
28. Linear: y(x)=x−1(C+e2x)
29. Linear: y(x)=(x2+x+C)(2x+1)−1/2
30. Substitution: v=x+y; solution: x=2(x+y)1/2−2 ln[1+(x+y)1/2]+C
31. Separable and linear
32. Separable and Bernoulli
33. Exact and homogeneous
34. Exact and homogeneous
35. Separable and linear
36. Separable and Bernoulli

Section 2.1

1. x(t)=22−e−t
2. x(t)=101+9e−10t
3. x(t)=2+e−2t2−e−2t
4. x(t)=3(1−e−12t)2(1+e−12t)
5. x(t)=408−3e−15t
6. x(t)=102+3e15t
7. x(t)=7711−4e−28t
8. x(t)=22117−4e91t
9. 484
10. 20 weeks
11. (b) P(t)=(12t+10)2
12. P(t)=24020−t
13. P(t)=18030−t
14. P(t)=P01+kP0t
16. About 27.69 months
17. About 44.22 months
19. About 24.41 months
20. About 42.12 months
21. 2001+e−6/5≈153.7 million
22. About 34.66 days
23. (a) limt→∞ x(t)=200 grams (b) 54ln 3≈1.37 seconds
24. About 9.24 days
25. (a) M=100 and k=0.0002; (b) In the year 2075
26. 50 ln 98≈15.89 months
27. (a) 100 ln95≈58.78 months; (b) 100 ln 2≈69.31 months.
28. (a) The alligators eventually die out.(b) Doomsday occurs after about 9 years 2 months.
29. (a) P(140)≈127.008 million (b) About 210.544 million; (c) In 2000 we get P≈196.169, whereas the actual 2000 population was about 281.422 million.
31. a≈0.3915; 2.15×106 cells
37. k≈0.0000668717, M≈338.027
38. k≈0.000146679, M≈208.250
39. P(t)=P0exp(kt+b2πsin 2 πt); the colored curve in the figure below shows the graph with P0=100, k=0.03, and b=0.06. It oscillates about the black curve which represents natural growth with P0=100 and k=0.03. We see that the two agree at the end of each full year.

Section 2.2

1. Unstable critical point: x=4;

x(t)=4+(x0−4)et
2. Stable critical point: x=3;

x(t)=3+(x0−3)e−t
3. Stable critical point: x=0; unstable critical point: x=4;

x(t) = 4x0x0+(4−x0)e4t
4. Stable critical point: x=3; unstable critical point: x=0;

x(t) = 3x0x0+(3−x0)e−3t
5. Stable critical point: x=−2; unstable critical point: x=2;

x(t) = 2[x0+2+(x0−2)e4t]x0+2+(x0−2)e4t
6. Stable critical point: x=3; unstable critical point: x=−3;

x(t) = 3[x0−3+(x0+3)e6t]x0−3+(x0+3)e4t
7. Semi-stable (see Problem 18 ) critical point: x=2;

x(t) = (2t−1)x0−4ttx0−2t−1
8. Semi-stable critical point: x=3;

x(t) = (3t+1)x0−9ttx0−3t+1
9. Stable critical point: x=1; unstable critical point: x=4;

x(t) = 4(1−x0)+(x0−4)e3t(1−x0)+(x0−4)e3t
10. Stable critical point: x=5; unstable critical point: x=2;

x(t) =2(5−x0)+5(x0−2)e3t(5−x0)+(x0−2)e3t
11. Unstable critical point: x=1;

1(x(t)−2)2=1(x0−2)2−2t
12. Stable critical point: x=2;

1(2−x(t))2=1(2−x0)2+2t

For each of Problems 13 through 18 we show a plot of slope field and typical solution curves. The equilibrium solutions of the given differential equation are labeled, and the stability or instability of each should be clear from the picture.

13.
14.
15.
16.
17.
18.
19. There are two critical points if h<212, one critical point if h=212, and no critical points if h>212. The bifurcation diagram is the parabola (c−5)2=25−10h in the hc-plane.
20. There are two critical points if s<116, one critical point if s=116, and no critical points if s>116. The bifurcation diagram is the parabola (2c−5)2=25(1−16s) in the sc-plane.

Section 2.3

1. Approximately 31.5 s
3. 400/(ln 2)≈577 ft
5. 400 ln 7≈778 ft
7. (a) 100 ft/sec; (b) about 23 sec and 1403 ft to reach 90 ft/sec
8. (a) 100 ft/sec; (b) about 14.7 sec and 830 ft to reach 90 ft/sec
9. 50 ft/s
10. About 5 min 47 s
11. Time of fall: about 12.5 s
12. Approximately 648 ft
19. Approximately 30.46 ft/s; exactly 40 ft/s
20. Approximately 277.26 ft
22. Approximately 20.67 ft/s; about 484.57 s
23. Approximately 259.304 s
24. (a) About 0:88 cm; (b) about 2:91 km
25. (b) About 1.389 km/sec; (c) rmax=100R/19≈5.26R
26. Yes
28. (b) After about 812 minutes it hits the surface at about 4.116 km/sec.
29. About 51.427 km
30. Approximately 11.11 km/sec (as compared with the earth’s escape velocity of about 11.18 km/sec).

Section 2.4

In Problems 1 through 10 we round off the indicated values to 3 decimal places.

1. Approximate values 1.125 and 1.181; true value 1.213
2. Approximate values 1.125 and 1.244; true value 1.359
3. Approximate values 2.125 and 2.221; true value 2.297
4. Approximate values 0.625 and 0.681; true value 0.713
5. Approximate values 0.938 and 0.889; true value 0.851
6. Approximate values 1.750 and 1.627; true value 1.558
7. Approximate values 2.859 and 2.737; true value 2.647
8. Approximate values 0.445 and 0.420; true value 0.405
9. Approximate values 1.267 and 1.278; true value 1.287
10. Approximate values 1.125 and 1.231; true value 1.333

Problems 11 through 24 call for tables of values that would occupy too much space for inclusion here. In Problems 11 through 16 we give first the final x-value, next the corresponding approximate x-values obtained with step sizes h=0.01 and h=0.005, and then the final true y-value. (All y-values rounded off accurate to 4 decimal places.)

11. 1.0, −0.7048, −0.7115, −0.7183
12. 1.0, 2.9864, 2.9931, 3.0000
13. 2.0, 4.8890, 4.8940, 4.8990
14. 2.0, 3.2031, 3.2304, 3.2589
15. 3.0, 3.4422, 3.4433, 3.4444
16. 3.0, 8.8440, 8.8445, 8.8451

In Problems 17 through 24 we give first the final x-value and then the corresponding approximate y-values obtained with step sizes h=0.1, h=0.02, h=0.004, and h=0.0008 respectively. (All y-values rounded off accurate to 4 decimal places.)

17. 1.0, 0.2925, 0.3379, 0.3477, 0.3497
18. 2.0, 1.6680, 1.6771, 1.6790, 1.6794
19. 2.0, 6.1831, 6.3653, 6.4022, 6.4096
20. 2.0, −1.3792, −1.2843, −1.2649, −1.2610
21. 2.0, 2.8508, 2.8681, 2.8716, 2.8723
22. 2.0, 6.9879, 7.2601, 7.3154, 7.3264
23. 1.0, 1.2262, 1.2300, 1.2306, 1.2307
24. 1.0, 0.9585, 0.9918, 0.9984, 0.9997
25. With both step sizes h=0.01 and h=0.005, the approximate velocity after 1 second is 16.0 ft/sec (80% of the limiting velocity of 20 ft/sec); after 2 seconds it is 19.2 ft/sec (96% of the limiting velocity).
26. With both step sizes h=1 and h=0.5, the approximate population after 5 years is 49 deer (65% of the limiting population of 75 deer); after 10 years it is 66 deer (88% of the limiting population).
27. With successive step sizes h=0.1, 0.01, 0.001,… the first four approximations to y(2) we obtain are 0.7772, 0.9777, 1.0017, and 1.0042. It therefore seems likely that y(2)≈1.00.
28. With successive step sizes h=0.1, 0.01, 0.001,… the first four approximations to y(2) we obtain are 1.2900, 1.4435, 1.4613, and 1.4631. It therefore seems likely that y(2)≈1.46.

29.

`x`	h=0.15 `y`	h=0.03 `y`	h=0.006 `y`
−1.0	1.0000	1.0000	1.0000
−0.7	1.0472	1.0512	1.0521
−0.4	1.1213	1.1358	1.1390
−0.1	1.2826	1.3612	1.3835
0.2	0.8900	1.4711	0.8210
0.5	0.7460	1.2808	0.7192

30.

x h=0.1
y h=0.01
y

1.8 2.8200 4.3308

1.9 3.9393 7.9425

2.0 5.8521 28.3926
31.

x h=0.1
y h=0.01
y

0.7 4.3460 6.4643

0.8 5.8670 11.8425

0.9 8.3349 39.5010

`x`	h=0.1 `y`	h=0.01 `y`
1.8	2.8200	4.3308
1.9	3.9393	7.9425
2.0	5.8521	28.3926

`x`	h=0.1 `y`	h=0.01 `y`
0.7	4.3460	6.4643
0.8	5.8670	11.8425
0.9	8.3349	39.5010

Section 2.5

1.

x Improved Euler y Actual y

0.1 1.8100 1.8097

0.2 1.6381 1.6375

0.3 1.4824 1.4816

0.4 1.3416 1.3406

0.5 1.2142 1.2131

`x`	Improved Euler `y`	Actual `y`
0.1	1.8100	1.8097
0.2	1.6381	1.6375
0.3	1.4824	1.4816
0.4	1.3416	1.3406
0.5	1.2142	1.2131

Note: In Problems 2 through 10, we give the value of x, the corresponding improved Euler value of y, and the true value of y.

2. 0.5, 1.3514, 1.3191
3. 0.5, 2.2949, 2.2974
4. 0.5, 0.7142, 0.7131
5. 0.5, 0.8526, 0.8513
6. 0.5, 1.5575, 1.5576
7. 0.5, 2.6405, 2.6475
8. 0.5, 0.4053, 0.4055
9. 0.5, 1.2873, 1.2874
10. 0.5, 1.3309, 1.3333

In Problems 11 through 16 we give the final value of x, the corresponding values of y with h=0.01 and with h=0.005, and the true value of y

11. 1.0, −0.71824, −0.71827, −0.71828
12. 1.0, 2.99995, 2.99999, 3.00000
13. 2.0, 4.89901, 4.89899, 4.89898
14. 2.0, 3.25847, 3.25878, 3.25889
15. 3.0, 3.44445, 3.44445, 3.44444
16. 3.0, 8.84511, 8.84509, 8.84509

In Problems 17 through 24 we give the final value of x and the corresponding values of y for h=0.1, 0.02, 0.004, and 0:0008.

17. 1.0, 0.35183, 0.35030, 0.35023, 0.35023
18. 2.0, 1.68043, 1.67949, 1.67946, 1.67946
19. 2.0, 6.40834, 6.41134, 6.41147, 6.41147
20. 2.0, −1.26092, −1.26003, −1.25999, −1.25999
21. 2.0, 2.87204, 2.87245, 2.87247, 2.87247
22. 2.0, 7.31578, 7.32841, 7.32916, 7.32920
23. 1.0, 1.22967, 1.23069, 1.23073, 1.23073
24. 1.0, 1.00006, 1.00000, 1.00000, 1.00000
25. With both step sizes h=0.01 and h=0.005 the approximate velocity after 1 second is 15.962 ft/sec (80% of the limiting velocity of 20 ft/sec); after 2 seconds it is 19.185 ft/sec (96% of the limiting velocity).
26. With both step sizes h=1 and h=0.5 the approximate population after 5 years is 49.391 deer (65% of the limiting population of 75 deer); after 10 years it is 66.113 deer (88% of the limiting population).
27. With successive step sizes h=0.1, 0.01, 0.001,… the first three approximations to y(2) we obtain are 1.0109, 1.0045, and 1.0045. It therefore seems likely that y(2)≈1.0045.
28. With successive step sizes h=0.1, 0.01, 0.001,… the first four approximations to y(2) we obtain are 1.4662, 1.4634, 1.4633, and 1.4633. It therefore seems likely that y(2)≈1.4633.
29. Impact speed approximately 43.22 m/s
30. Impact speed approximately 43.48 m/s

Section 2.6

1. y(0.25)≈1.55762; y(0.25)=1.55760. y(0.5)≈1.21309; y(0.5)=1.21306. Solution: y=2e−x

In Problems 2 through 10 we give the approximation to y(0.5), its true value, and the solution.

2. 1.35867, 1.35914; y=12e2x
3. 2.29740, 2.29744; y=2ex−1
4. 0.71309, 0.71306; y=2e−x+x−1
5. 0.85130, 0.85128; y=−ex+x+2
6. 1.55759, 1.55760; u=2 exp(−x2)
7. 2.64745, 2.64749; y=3 exp(−x3)
8. 0.40547, 0.40547; y=ln(x+1)
9. 1.28743, 1.28743; y=tan14(x+π)
10. 1.33337, 1.33333; y=(1−x2)−1

11. Solution: y(x)=2−ex.

`x`	h=0.2 `y`	h=0.1 `y`	Exact `y`
0.0	1.00000	1.00000	1.00000
0.2	0.77860	0.77860	0.77860
0.4	0.50818	0.50818	0.50818
0.6	0.17789	0.17788	0.17788
0.8	−0.22552	−0.22554	−0.22554
1.0	−0.71825	−0.71828	−0.71828

In Problems 12 through 16 we give the final value of x, the corresponding Runge-Kutta approximations with h=0.2 and with h=0.1, the exact value of y, and the solution.

12. 1.0, 2.99996, 3.00000, 3.00000; y=1+2/(2−x)
13. 2.0, 4.89900, 4.89898, 4.89898; y=8+x4
14. 2.0, 3.25795, 3.25882, 3.25889; y=1/(1−ln x)
15. 3.0, 3.44445, 3.44444, 3.44444; y=x+4x−2
16. 3.0, 8.84515, 8.84509, 8.84509; y=(x6−37)1/3

In Problems 17 through 24 we give the final value of x and the corresponding values of y with h=0.2, 0.1, 0.05, and 0.025.

17. 1.0, 0.350258, 0.350234, 0.350232, 0.350232
18. 2.0, 1.679513, 1.679461, 1.679459, 1.679459
19. 2.0, 6.411464, 6.411474, 6.411474, 6.411474
20. 2.0, −1.259990, −1.259992, −1.259993, −1.259993
21. 2.0, 2.872467, 2.872468, 2.872468, 2.872468
22. 2.0, 7.326761, 7.328452, 7.328971, 7.329134
23. 1.0, 1.230735, 1.230731, 1.230731, 1.230731
24. 1.0, 1.000000, 1.000000, 1.000000, 1.000000
25. With both step sizes h=0.1 and h=0.05, the approximate velocity after 1 second is 15.962 ft/sec (80% of the limiting velocity of 20 ft/sec); after 2 seconds it is 19.185 ft/sec (96% of the limiting velocity).
26. With both step sizes h=6 and h=3, the approximate population after 5 years is 49.3915 deer (65% of the limiting population of 75 deer); after 10 years it is 66.1136 deer (88% of the limiting population).
27. With successive step sizes h=1, 0.1, 0.01,… the first four approximations to y(2) we obtain are 1.05722, 1.00447, 1.00445and 1.00445. Thus it seems likely that y(2)≈1.00445 accurate to 5 decimal places.
28. With successive step sizes h=1, 0.1, 0.01,… the first four approximations to y(2) we obtain are 1.48990, 1.46332, 1.46331, and 1.46331. Thus it seems likely that y(2)≈1.4633 accurate to 5 decimal places.
29. Time aloft: approximately 9.41 seconds
30. Time aloft: approximately 9.41 seconds

Section 3.1

1. x=3, y=2
2. x=5, y=−3
3. x=−4, y=3
4. x=5, y=4
5. Inconsistent—no solution
6. Inconsistent—no solution
7. x=−10+4t, y=t (infinitely many solutions)
8. x=4+2t, y=t (infinitely many solutions)
9. x=4, y=−1, z=3
10. x=3, y=1, z=−2
11. x=1, y=3, z=−4
12. x=1, y=3, z=5
13. x=y=z=0
14. x=5, y=3, z=−4
15. Inconsistent—no solution
16. Inconsistent—no solution
17. Inconsistent—no solution
18. Inconsistent—no solution
19. x=8+3t, y=3+2t, z=t (infinitely many solutions)
20. x=−5t, y=5+t, z=t (infinitely many solutions)
21. x=3−2t, y=2+3t, z=t (infinitely many solutions)
22. x=−4t, y=−5t, z=t (infinitely many solutions)
23. y(x)=3 cos 2x+4 sin 2x
24. y(x)=5 cosh3x+4 sinh3x
25. y(x)=7e5x+3e−5x
26. y(x)=23e11x+21e−11x
27. y(x)=23e3x+17e−5x
28. y(x)=23e3x−8e7x
29. y(x)=52ex/2−45ex/3
30. y(x)=81e4x/3−40e−7x/5
31. Suggestion: The two lines both pass through the origin.
32. Suggestion: Two distinct planes in space either are parallel or intersect in a straight line.
33. (a) No solution (b) A unique solution (c) No solution (d) No solution (e) A unique solution (f) Infinitely many solutions
34. (a) No solution (b) Infinitely many solutions (c) No solution (d) No solution (e) Infinitely many solutions (f) A unique solution

Section 3.2

1. x1=1, x2=0, x3=2
2. x1=5, x2=1, x3=−3
3. x1=13+11t, x2=2+5t, x3=t
4. x1=35+33t, x2=5+7t, x3=t
5. x1=13+4t, x2=6+t, x3=5+3t, x4=t
6. x1=17+t, x2=11+3t, x3=t, x4=−4
7. x1=3−8s+19t, x2=7+2s−7t, x3=s, x4=t
8. x1=−25+10s+22t, x2=s, x3=10−3t, x4=t
9. x1=1, x2=3, x3=−5, x4=6
10. x1=63s−16t, x2=13s−8t, x3=s, x4=5t, x5=t
11. x1=3, x2=−2, x3=4
12. x1=5, x2=−3, x3=2
13. x1=4+3t, x2=3−2t, x3=t
14. x1=5+2t, x2=t, x3=7
15. Inconsistent—no solution
16. Inconsistent—no solution
17. x1=3−2t, x2=−4+t, x3=5−3t, x4=t
18. x1=4+2s−3t, x2=s, x3=3−4t, x4=t
19. x1=3−s−t, x2=5+2s−3t, x3=s, x4=t
20. x1=2+3t, x2=1+s−2t, x3=2+2s, x4=s, x5=t
21. x1=2, x2=1, x3=3, x4=4
22. x1=3, x2=−2, x3=4, x4=−1
23. (a) None (b) k≠2 (c) k=2
24. (a) k≠2 (b) None (c) k=4
25. (a) k≠4 (b) k=4 (c) None
26. (a) All k (b) None (c) None
27. (a) None (b) k≠11 (c) k=11
28. No solution unless c=2a+3b, in which case it has infinitely many solutions.

Section 3.3

1. [1001]
2. [1001]
3. [10−2013]
4. [10201−1]
5. [10−501−3]
6. [10701−6]
7. [10501−1000]
8. [100010001]
9. [102014000]
10. [10−3015000]
11. [130001000]
12. [1−40001000]
13. [1003010−20012]
14. [1004010−30015]
15. [10−12013−10000]
16. [103−201430000]
17. [1002−3010−14001−2−5]
18. [103−230145100000]
19. [1021301−2−3100000]
20. [120230011400000]
21. x1=3, x2=−2, x3=4
22. x1=5, x2=−3, x3=2
23. x1=4+3t, x2=3−2t, x3=t
24. x1=5+2t, x2=t, x3=7
25. Inconsistent—no solution
26. Inconsistent—no solution
27. x1=3−2t, x2=−4+t, x3=5−3t, x4=t
28. x1=4+2s−3t, x2=s, x3=3−4t, x4=t
29. x1=3−s−t, x2=5+2s−3t, x3=s, x4=t
30. x1=2+3t, x2=1+s−2t, x3=2+2s, x4=s, x5=t
31. The sequence 16R3, R2−5R3, 14R2, R1−2R2, R1−3R3 of row operations transforms the first matrix to the identity matrix.
33. [1001], [1*00], [0100], [0000]

Section 3.4

1. [5−15185]
2. [16−9−18−262215]
3. [−262012−62218]
4. [44−22−205310−26352194]
5. AB=[−91−1012], BA=[−28115]
6. AB=[7−13−2423104111−857], BA=[1−17−221216727−2157]
7. AB=[26], BA=[369481251015]
8. AB=[2115350], BA=[3097−201316−2538]
9. BA=[47−22] but AB is not defined.
10. AB=[1−2135−631] but BA is not defined.
11. AB=[11153] but BA is not defined.
12. Neither product matrix AB nor BA is defined.
13. A(BC)=(AB)C=[3251−2−17]
14. A(BC)=(AB)C=[−3]
15. A(BC)=(AB)C=[1215810]
16. A(BC)=(AB)C=[−4−4−22−9−12−912−14−18−1317]
17. x=s(5, −2, 1, 0)+t(−4, 7, 0, 1)
18. x=s(3, 1, 0, 0)+t(−6, 0, −9, 1)
19. x=s(−3, 2, −1, 1, 0)+t(1, −6, 8, 0, 1)
20. x=s(3, 1, 0, 0, 0)+t(−7, 0, 2, 10, 0)
21. x=r(1, −2, 1, 0, 0)+s(−2, 3, 0, 1, 0 )+t(−7, −4, 2, 0, 1)
22. x=r(1, 1, 0, 0, 0)+s(−7, 0, 1, 1, 0 )+t(−3, 0, 2, 0, 1)
23. B=[2−1−32]
24. B=[7−4−53]
25. B=[3−7−25]
29. A2=[a2+bcab+bdac+dcbc+d2], A2+(ad−bc)I=[a2+adab+bdac+dcd2+ad]=[a(a+d)b(a+d)c(a+d)d(a+d)]=(a+d)A
31. (a) (A+B)(A−B)=[−25−34−71−34]≠[−8−45−44−51]=A2−B2
33. [1001], [−100−1], [+100−1], [−100+1]
35. [2−12−1]
37. [01−10]
39. A(c1x1+c2x2)=c1Ax1+c2Ax2=c1 ·0+c2 ·0=0
43. A2=[6−3−3−36−3−3−36]=3A

Section 3.5

1. A−1=[3−2−43], x=[3−2]
2. A−1=[5−7−23], x=[−2611]
3. A−1=[6−7−56], x=[33−28]
4. A−1=[17−12175], x=[25−10]
5. A−1=12[4−2−53], x=12[8−7]
6. A−1=13[6−7−34], x=13[25−10]
7. A−1=14[7−9−57], x=14[3−1]
8. A−1=15[10−15−58], x=15[25−11]
9. A−1=[5−6−45]
10. A−1=12[6−7−45]
11. A−1=[−5−2521−2−4−35]
12. A−1=[182−7−301−4−12]
13. A−1=[−1342−53−912−71]
14. A−1=[11−7−9−433−212]
15. A−1=[−2227−273810−1−3]
16. A−1=13[−303−1−3−1−132]
17. A−1=14[−2−6−3−2−2−1−2−21]
18. A−1=15[12−2−505−3−16]
19. A−1=16[−211189−5−2−330]
20. A−1=17[310−2−37−120]
21. A−1=[0100−201010000−301]
22. A−1=[1−1100−2−12011−1−33−51]
23. A−1=[4−3−54], X=[718−35−9−2345]
24. A−1=[7−6−87], X=[14−3046−1635−53]
25. A−1=[11−94−22−1−210], X=[7−1415−13−2−22−4]
26. A−1=[−163116−1−4−132−9], X=[−21968−3−2−1765]
27. A−1=[7−20170−11−26−5], X=[17−2024−131−11−1−56−74]
28. A−1=[−5510−881524−23−45], X=[−55101−8815724−23−45−13]

Section 3.6

1. 60
2. 4
3. −210
4. −72
5. 120
6. 60
7. 0
8. 25
9. 30
10. 7
11. 40
12. 10
13. 78
14. −22
15. −74
16. 84
17. 8
18. 135
19. 39
20. 79
21. Δ=1, x=10, y=−7
22. Δ=1, x=−1, y=1
23. Δ=1, x=2, y=−4
24. Δ=1, x=5, y=−3
25. Δ=2, x=6, y=−3
26. Δ=−2, x=12, y=0
27. Δ=96, x1=13, x2=−23, x3=−13
28. Δ=35, x1=47, x2=37, x3=27
29. Δ=23, x1=2, x2=3, x3=0
30. Δ=56, x1=−17, x2=914, x3=27
31. Δ=14, x1=−87, x2=−107, x3=17
32. Δ=6, x1=−73, x2=9, x3=8
33. det A=−4, A−1=14[444161513282523]
34. det A=35, A−1=14[−2−3129−4−19132−8]
35. det A=35, A−1=135[−1525−2610−5815−2519]
36. det A=23, A−1=123[520−171017−1114−8]
37. det A=29, A−1=129[11−14−15−17191018−15−14]
38. det A=6, A−1=16[−610215−21612−18−6]
39. det A=37, A−1=137[−21−1−13496−65−9]
40. det A=107, A−1=1107[912−1311−21−4−15−20−14]

Section 3.7

1. y(x)=−2+3x
2. y(x)=4−7x
3. y(x)=3−2x2
4. y(x)=2x+3x2
5. y(x)=5−3x+x2
6. y(x)=−10−7x+2x2
7. y(x)=13(4x+3x2−4x3)
8. y(x)=5+3x−x3
9. y(x)=4+3x+2x2+x3
10. y(x)=17−5x+3x2−2x3
11. x2+y2−6x−4y−12=0, center (3, 2) and radius 5
12. x2+y2+6x−8y−75=0, center (−3, 4) and radius 10
13. x2+y2+4x+4y−5=0, center (−2,−2) and radius 13
14. x2+y2−10x−24y=0, center (5, 12) and radius 13
15. x2−xy+y2=25
16. 4x2−7xy+4y2=100
17. 100x2−199xy+100y2=100
18. 400x2−481xy+225y2=3600
19. y=3+2x
20. y=10+8x−16x2
21. x2+y2+z2−2x−4y−6z−155=0, center (1, 2, 3) and radius 13
22. x2+y2+z2−10x+14y−18z−521=0, center (5, −7, 9) and radius 26
23. P(t)=49.061−0.0722t+0.00798t2
24. P(t)=56.590+0.30145t−0.007375t2
25. P(t)=62.813+1.37915t−0.012375t2
26. P(t)=34.838+0.7692t+0.0064t2
27. P(t)=44.678+0.850417t−0.05105t2+0.000983833t3
28. P(t)=51.619+0.672433t−0.019565t2+0.000203167t3
29. P(t)=54.973+0.308667t+0.059515t2−0.00119817t3
30. P(t)=28.053+0.592233t+0.00907t2−0.0000443333t3
31. P(t)=39.478+0.209692t+0.0564163t2−0.00292992t3+0.0000391375t4
32. P(t)=44.461+0.7651t−0.000489167t2−0.000516t3+7.19167×10−6t4
33. P(t)=47.197+1.22537t−0.077192t2+0.00373475t3−0.0000493292t4
34. P(t)=20.190+1.00003t−0.031775t2+0.00116067t3−0.000041205t4
36. y=2x2−6x+7
38. x2+y2+6x−8y−75=0, center (−3, 4) and radius 10
40. 400x2−481xy+225y2−3600=0

Section 4.1

1. 51, (5, 8, −11), (2, 23, 0)
2. 9, (1, 4, −1), (−15, −16, 26)
3. 321, 9i−3j+3k, −14i−21j+43k
4. 17, 4i−j−3k, 6i−7j+12k
5. Linearly dependent
6. Linearly dependent
7. Linearly dependent
8. Linearly dependent
9. w=3u+2v
10. w=2u−3v
11. w=u−2v
12. w=3u+5v
13. w=2u−3v
14. w=7u+5v
15. Linearly dependent
16. Linearly dependent
17. Linearly dependent
18. Linearly dependent
19. Linearly dependent; a=3, b=2, c=1
20. Linearly dependent; a=−2, b=3, c=1
21. Linearly dependent; a=11, b=4, c=−1
22. Linearly dependent
23. Linearly dependent
24. Linearly dependent
25. t=2u−v+3w
26. t=u+5v−w
27. t =2u+6v+w
28. t =u+v+w

Section 4.2

1. W is a subspace of R3.
2. W is a subspace of R3.
3. W is not a subspace of R3.
4. W is not a subspace of R3.
5. W is a subspace of R4.
6. W is a subspace of R4.
7. W is not a subspace of R2.
8. W is a subspace of R2.
9. W is not a subspace of R2.
10. W is not a subspace of R2.
11. W is a subspace of R4.
12. W is not a subspace of R4.
13. W is not a subspace of R4.
14. W is not a subspace of R4.
15. u=(−1, 0, 1, 0) and v=(−4, −2, 0, 1)
16. u=(−1, −1, 0, 1) and v=(−5, −3, 0, 1)
17. u=(1, −3, −2, 0) and v=(−2, 1, 0, 1)
18. u=(2, −1, −2, 1, 0) and v=(3, −4, 5, 0, 1)
19. u=(1, 2, −1, 0)
20. u=(−5, 3, −2, 1)
21. u=(−3, 2,−4, 1)
22. u=(−6, 4,−3, 1)

Section 4.3

1. Linearly dependent
2. Linearly dependent
3. Linearly dependent
4. Linearly dependent
5. Linearly dependent
6. Linearly dependent
7. Linearly dependent
8. Linearly dependent
9. w=2v1−3v2
10. w=7v1+4v2
11. w=v1−2v2
12. w=2v1+5v2
13. w cannot be expressed as a linear combination of v1 and v2
14. w cannot be expressed as a linear combination of v1, v2 and v3
15. w=3v1−2v2+4v3
16. w=6v1−2v2+3v3
17. The vectors v1, v2, v3 are linearly independent.
18. 3v1+v2+ 5v3=0
19. The vectors v1, v2, v3 are linearly independent.
20. The vectors v1, v2, v3 are linearly independent.
21. v1−2v2− v3=0
22. 7v1+5v2− 9v3=0

Section 4.4

1. The vectors v1 and v2 form a basis for R2.
2. The vectors v1, v2, v3 do not form a basis for R3.
3. The given vectors do not form a basis for R3.
4. The given vectors do not form a basis for R4.
5. The three vectors v1, v2, v3 do not form a basis for R3.
6. The four given vectors form a basis for R3.
7. The three given vectors form a basis for R3.
8. The three given vectors form a basis for R4.
9. The plane x−2y+5z=0 is a 2-dimensional subspace of R3 with basis consisting of the vectors v1=(2, 1, 0) and v2=(−5, 0, 1).
10. The plane y−z=0 is a 2-dimensional subspace of R3 with basis consisting of the vectors v1=(1, 0, 0) and v2=(0, 1, 1).
11. The line is a 1-dimensional subspace of R3 with basis consisting of the vector v=(−3, 1, 1).
12. Hence the subspace consisting of all such vectors is 3-dimensional with basis consisting of the vectors v1‘=(1, 1, 0, 0), v2=(1, 0, 1, 0), and v3=(1, 0, 0, 1).
13. The subspace consisting of all such vectors is 2-dimensional with basis consisting of the vectors v1=(3, 0, 1, 0) and v2=(0, 4, 0, 1).
14. The subspace consisting of all such vectors is 2-dimensional with basis consisting of the vectors v1=(−2, 1, 0, 0) and v2=(0, 0,−3, 1).
15. The solution space of the given system is 1-dimensional with basis consisting of the vector v1=(11, 7, 1).
16. The solution space of the given system is 1-dimensional with basis consisting of the vector v1=(11,−5, 1).
17. The solution space of the given system is 2-dimensional with basis consisting of the vectors v1=(−11,−3, 1, 0) and v2(−11,−5, 0, 1).
18. The solution space of the given system is 2-dimensional with basis consisting of the vectors v1=(−3, 1, 0, 0) and v2(−25, 0, 5, 1).
19. The solution space of the given system is 2-dimensional with basis consisting of the vectors v1=(3,−2, 1, 0) and v2=(−4,−3, 0, 1).
20. The solution space of the given system is 2-dimensional with basis consisting of the vectors v1=(1,−3, 1, 0) and v2=(−2, 1, 0, 1).
21. The solution space of the given system is 2-dimensional with basis consisting of the vectors v1=(−1,−1, 1, 0) and v2=(−5,−3, 0, 1).
22. The solution space of the given system is 2-dimensional with basis consisting of the vectors v1=(2, 1, 0, 0) and v2=(−5, 0,−7, 1).
23. The solution space of the given system is 1-dimensional with basis consisting of the vector v1=(2,−3, 1, 0).
24. The solution space of the given system is 3-dimensional with basis consisting of the vectors v1=(−2, 2, 1, 0, 0), v2=(−1, 3, 0, 1, 0), and v3=(−3,−1, 0, 0, 1).
25. The solution space of the given system is 3-dimensional with basis consisting of the vectors v1=(−2, 1, 0, 0, 0), v2=(2, 0, 1, 1, 0), and v3=(−3, 0,−4, 0, 1).
26. The solution space of the given system is 2-dimensional with basis consisting of the vectors v1=(−2, 1, 2, 1, 0) and v2=(3,−4, 5, 0, 1).

Section 4.5

1. Row basis: The first and second row vectors of E. Column basis: The first and second column vectors of A.
2. Row basis: The first and second row vectors of E. Column basis: The first and second column vectors of A.
3. Row basis: The first and second row vectors of E. Column basis: The first and second column vectors of A.
4. Row basis: The three row vectors of E. Column basis: The first three column vectors of A.
5. Row basis: The three row vectors of E. Column basis: The first, second, and fourth column vectors of A.
6. Row basis: The three row vectors of E. Column basis: The first, second, and fourth column vectors of A.
7. Row basis: The first two row vectors of E. Column basis: The first two column vectors of A.
8. Row basis: The first three row vectors of E. Column basis: The first, second, and fourth column vectors of A.
9. Row basis: The first three row vectors of E. Column basis: The first three column vectors of A.
10. Row basis: The first three row vectors of E. Column basis: The first, second, and fourth column vectors of A.
11. Row basis: The first three row vectors of E. Column basis: The first, second, and fifth column vectors of A.
12. Row basis: The first three row vectors of E. Column basis: The first, second, and fifth column vectors of A.
13. Linearly independent: v1 and v2
14. Linearly independent: v1 and v2
15. Linearly independent: v1, v2, and v4
16. Linearly independent: v1, v2, v4, and v5
17. Basis vectors: v1, v2, e2
18. Basis vectors: v1, v2, e2
19. Basis vectors: v1, v2, e2, e3
20. Basis vectors: v1, v2, e1, e3
21. The first and second equations are irredundant.
22. The first and second equations are irredundant.
23. The first, second, and fourth equations are irredundant.
24. The first, second, and fifth equations are irredundant.

Section 4.6

1. Yes, the three vectors are mutually orthogonal.
2. Yes, the three vectors are mutually orthogonal.
3. Yes, the three vectors are mutually orthogonal.
4. Yes, the three vectors are mutually orthogonal.
5. a2=7, b2=7, c2=14
6. a2=18, b2=18, c2=36
7. a2=19, b2=25, c2=44
8. a2=103, b2=112, c2=215
9. ∠A=∠B=45°
10. ∠A=∠B=45°
11. ∠A=41.08°, ∠B=48.92°
12. ∠A=43.80°, ∠B=46.20°
13. u1=(2, 1, 0), u2=(−3, 0, 1)
14. u1=(−5, 1, 0), u2=(3, 0, 1)
15. u1=(2, 1, 0, 0), u2=(3, 0, 1, 0), u3=(−5, 0, 0, 1)
16. u1=(−7, 1, 0, 0), u2=(6, 0, 1, 0), u3=(9, 0, 0, 1)
17. u1=(7,−3, 1, 0), u2=(−19, 5, 0, 1)
18. u1=(−12,−3, 1, 0), u2=(16, 7, 0, 1)
19. u1=(−13, 4, 1, 0, 0), u2=(4,−3, 0, 1, 0), u3=(−11, 4, 0, 0, 1)
20. u1=(−5, 1, 1, 0, 0), u2=(−12, 4, 0, 1, 0), u3=(−19, 7, 0, 0, 1)
21. u1=(−1,−1, 1, 0, 0), u2=(0,−1, 0,−1, 1)
22. u1=(−2, 1, 1, 0, 0), u2=(1,−2, 0, 1, 0)

Section 4.7

1. It is a subspace.
2. It is a subspace.
3. It is a not subspace.
4. It is a not subspace.
5. It is a subspace.
6. It is a not subspace.
7. It is a not subspace.
8. It is a subspace.
9. It is a not subspace.
10. It is a subspace.
11. It is a subspace.
12. It is a not subspace.
13. The functions sin x and cos x are linearly independent.
14. The funtions ex and xex are linearly independent.
15. The functions (1+x), (1−x), and (1−x2) are linearly independent.
16. The three given polynomials are linearly dependent.
17. The three given trigonometric functions are linearly dependent.
18. The two given linear combinations of sin x and cos x are linearly independent.
19. A=3 and B=−2.
20. A=−2 and B=C=1.
21. A=2, B=−2, and C=0.
22. A=−1, B=4, and C=−3.
23. The solution space is 3-dimensional with basis {1, x, x2}.
24. The solution space is 4-dimensional with basis {1, x, x2, x3}.
25. The solution space is 2-dimensional with basis {1, e5x}.
26. The solution space is 2-dimensional with basis {1, e−10x}.

Section 5.1

1. y(x)=52ex−52e−x
2. y(x)=2e3x−3e−3x
3. y(x)=3 cos 2x+4 sin 2x
4. y(x)=10 cos 5x−2 sin 5x
5. y(x)=2ex−e2x
6. y(x)=4e2x+3e−3x
7. y(x)=6−8e−x
8. y(x)=13(14−2e3x)
9. y(x)=2e−x+xe−x
10. y(x)=3e5x−2xe5x
11. y(x)=5exsin x
12. y(x)=e−3x(2 cos 2x+3 sin 2x)
13. y(x)=5x−2x2
14. y(x)=3x2−16/x3
15. y(x)=7x−5xln x
16. y(x)=2 cos(ln x)+3 sin(ln x)
21. Linearly independent
22. Linearly independent
23. Linearly independent
25. Linearly dependent
25. Linearly independent
26. Linearly independent
28. y(x)=1−2 cos x−sin x
29. There is no contradiction because if the given differential equation is divided by x2 to get the form in Eq. (8), then theresulting coefficient functions p(x)=−4/x and q(x)=6/x2 are not continuous at x=0.
33. y(x)=c1ex+c2e2x
34. y(x)=c1e−5x+c2e3x
35. y(x)=c1+c2e−5x
36. y(x)=c1+c2e−3x/2
37. y(x)=c1e−x/2+c2ex
38. y(x)=c1e−x/2+c2e−3x/2
39. y(x)=(c1+c2x)e−x/2
40. y(x)=(c1+c2x)e2x/3
41. y(x)=c1e−4x/3+c2e5x/2
42. y(x)=c1e−4x/7+c2e3x/5
43. y″+10y′=0
44. y″−100y=0
45. y″+20y′+100y=0
46. y″−110y′+1000y=0
47. y″=0
48. y″−2y′−y=0
49. The high point is (ln 74, 167).
50. (−ln 2, −2)
52. y(x)=c1x+c2/x
53. y(x)=c1x−4+c2x3
54. y(x)=c1x−3/2+c2x1/2
55. y(x)=c1+c2 ln x
56. y(x)=x2(c1+c2 ln x)

Section 5.2

1. 15·(2x)−16·(3x2)−6·(5x−8x2)≡0
2. (−4)(5)+(5)(2−3x2)+(1)(10+15x2)≡0
3. 1·0+0· sin x+0·ex≡0
4. (6)(17)+(−51)(2 sin2 x)+(−34)(3 cos2 x)≡0
5. 1·17−34·cos2 x+17·cos 2x≡0
6. (−1)(ex)+(1)(coshx)+(1)(sinhx)≡0
13. y(x)=43ex−13e−2x
14. y(x)=12(3ex−6e2x+3e3x)
15. y(x)=(2−2x+x2)ex
16. y(x)=−12ex+13e2x−10xe2x
17. y(x)=19(29−2 cos 3x−3 sin 3x)
18. y(x)=ex(2− cos x− sin x)
19. y(x)=x+2x2+3x3
20. y(x)=2x−x−2+x−2 ln x
21. y(x)=2 cos x−5 sin x+3x
22. y(x)=4e2x−e−2x−3
23. y(x)=e−x+4e3x−2
24. y(x)=ex(3 cos x+4 sin x)+x+1
38. y2(x)=1x3
39. y2(x)=xex/2
40. y2(x)=xex
41. y2(x)=x+2
42. y2(x)=1+x2

Section 5.3

1. y(x)=c1e2x+c2e−2x
2. y(x)=c1+c2e3x/2
3. y(x)=c1e2x+c2e−5x
4. y(x)=c1ex/2+c2e3x
5. y(x)=c1e−3x+c2xe−3x
6. y(x)=e−5x/2[c1exp(12x5)+c2exp(−12x5)]
7. y(x)=c1e3x/2+c2xe3x/2
8. y(x)=e3x(c1 cos 2x+c2 sin 2x)
9. y(x)=e−4x(c1 cos 3x+c2 sin 3x)
10. y(x)=c1+c2x+c3x2+c4e−3x/5
11. y(x)=c1+c2x+c3e4x+c4xe4x
12. y(x)=c1+c2ex+c3xex+c4x2ex
13. y(x)=c1+c2e−2x/3+c3xe−2x/3
14. y(x)=c1ex+c2e−x+c3 cos 2x+c4 sin 2x
15. y(x)=c1e2x+c2xe2x+c3e−2x+c4xe−2x
16. y(x)=(c1+c2x) cos 3x+(c3+c4x) sin 3x
17. y(x)=c1 cos (x/2)+c2 sin (x/2)+c3 cos (2x/3)+c4 sin (2x/3)
18. y(x)=c1e2x+c2e−2x+c3 cos 2x+c4 sin 2x
19. y(x)=c1ex+c2e−x+c3xe−x
20. y(x)=e−x/2[(c1+c2x) cos (12x3)+(c3+c3x) sin (12x3)]
21. y(x)=5ex+2e3x
22. y(x)=e−x/3[3 cos (x/3)+53 sin (x/3)]
23. y(x)=e3x(3 cos 4x−2 sin 4x)
24. y(x)=12(−7+e2x+8e−x/2)
25. y(x)=14(−13+6x+9e−2x/3)
26. y(x)=15(24−9e−5x−25xe−5x)
27. y(x)=c1ex+c2e−2x+c3xe−2x
28. y(x)=c1e2x+c2e−x+c3e−x/2
29. y(x)=c1e−3x+e3x/2[c2 cos (32x3)+c3 sin (32x3)]
30. y(x)=c1e−x+c2e2x+c3 cos (x3)+c4 sin (x3)
31. y(x)=c1ex+e−2x(c2 cos 2x+c3 sin 2x)
32. y(x)=c1e2x+(c2+c3x+c4x2)e−x
33. y(x)=c1e3x+e−3x(c2 cos 3x+c3 sin 3x)
34. y(x)=c1e2x/3+c2 cos 2x+c3 sin 2x
35. y(x)=c1e−x/2+c2e−x/3+c3 cos 2x+c4 sin 2x
36. y(x)=c1e7x/9+e−x(c2 cos x+c3 sin x)
37. y(x)=11+5x+3x2+7ex
38. y(x)=2e5x−2 cos 10x
39. y(3)−6y″+12y′−8y=0
40. y(3)−2y″+4y′−8y=0
41. y(4)−16y=0
42. y(6)+12y(4)+48y″+64y=0
44. (a) x=i, −2i (b) x=−i, 3i
45. y(x)=c1e−ix+c2e3ix
46. y(x)=c1e3ix+c2e−2ix
47. y(x)=c1exp([1+i3]x)+c2exp(−[1+i3]x)
48. y(x)=13(ex+exp[12(−1+i3)x]+exp[12(−1−i3)x])
49. y(x)=2e2x−5e−x+3 cos x−9 sin x
52. y(x)=c1 cos (3 ln x)+c2 sin (3 ln x)
53. y(x)=x−3[c1 cos (4 ln x)+c2 sin (4 ln x)]
54. y(x)=c1+c2 ln x+c3x−3
55. y(x)=c1+x2(c2+c3 ln x)
56. y(x)=c1+c2 ln x+c3(ln x)2
57. y(x)=c1+x3(c2x−3+c3x+3)
58. y(x)=x−1[c1+c2 ln x+c3(ln x)2]

Section 5.4

1. Frequency: 2 rad/s (1/π HZ); period: π s
2. Frequency: 8 rad/sec (4/π HZ); period: π/4 sec
3. Amplitude: 2m; frequency: 5 rad/s; period: 2π/5 s
4. (a) x(t)=1312cos (12t−α) with α=2π−tan−1(5/12)≈5.8884; (b) Amplitude: 1312 m; period: π/6 sec
6. About 7.33 mi
7. About 10450 ft
8. 29.59 in.
10. Amplitude: 100 cm; period: about 2.01 sec
11. About 3.8 in.
13. (a) x(t)=50(e−2t/5−e−t/2); (b) 4.096 exactly
14. (a) x(t)=25e−t/5 cos (3t−α) with α=tan−1(3/4)≈0.6435; (b) envelope curves x=±25e−t/5; pseudoperiod 2π/3
15. x(t)=4e−2t−2e−4t, overdamped; u(t)=2 cos (22 t)
16. x(t)=4e−3t−2e−7t, overdamped; u(t)≈22221 cos (21 t−0.2149)
17. x(t)=5e−4t(2t+1), critically damped; u(t)≈525 cos (4t−5.8195)
18. x(t)=2e−3t cos (4t−3π2), underdamped; u(t)=85cos (5t−3π2)
19. x(t)≈13313 e−5t/2 cos (6t−0.8254), underdamped; u(t)≈413233 cos (132t−0.5517)
20. x(t)≈13e−4t cos (2t−1.1760), underdamped; u(t)≈1295 cos (25 t−0.1770)
21. x(t)≈10e−5t cos (10t−0.9273), underdamped; u(t)≈214 cos (55 t−0.6405)
22. (b) The time-varying amplitude is 233, the frequencyis 43 rad/s, and the phase angle is π/6.
23. (a) k≈7018 lb/ft; (b) After about 2.47 s
34. Damping constant: c≈11.51 lb/ft/s; spring constant: k≈189.68 lb/ft

Section 5.5

1. yp(x)=125e3x
2. yp(x)=−14(5+6x)
3. yp(x)=139(cos 3x−5 sin 3x)
4. yp(x)=19(−4ex+3xex)
5. yp(x)=126(13+3 cos 2x−2 sin 2x)
6. yp(x)=1343(4−56x+49x2)
7. yp(x)=−16(ex−e−x)=−13sinhx
8. yp(x)=14xsinh2x
9. yp(x)=−13+116(2x2−x)ex
10. yp(x)=16(2x sin 3x−3x cos 3x)
11. yp(x)=18(3x2−2x)
12. yp(x)=2x+12x sin x
13. yp(x)=165ex(7 sin x−4 cos x)
14. yp(x)=124(−3x2ex+x3ex)
15. yp(x)≡−17
16. yp(x)=181(45+e3x−6xe3x+9x2e3x)
17. yp(x)=14(x2 sin x−x cos x)
18. yp(x)=−1144(24xex−19xe2x+6x2e2x)
19. yp(x)=18(10x2−4x3+x4)
20. yp(x)=−7+13xex
21. yp(x)=xex(A cos x+B sin x)
22. yp(x)=Ax3+Bx4+Cx5+Dxex
23. yp(x)=Ax cos 2x+Bx sin 2x+Cx2 cos 2x+Dx2 sin 2x
24. yp(x)=Ax+Bx2+(Cx+Dx2)e−3x
25. yp(x)=Axe−x+Bx2e−x+Cxe−2x+Dx2e−2x
26. yp(x)=(Ax+Bx2)e3x cos 2x+(Cx+Dx2)e3x sin 2x
27. yp(x)=Ax cos x+Bx sin x+Cx cos 2x+Dx sin 2x
28. yp(x)=(Ax+Bx2+Cx3) cos 3x+(Dx+Ex2+Fx3) sin 3x
29. yp(x)=Ax3ex+Bx4ex+Cxe2x+Dxe−2x
30. yp(x)=(A+Bx+Cx2) cos x+(D+Ex+Fx2) sin x
31. y(x)=cos 2x+34sin 2x+12x
32. y(x)=16(15e−x−16e−2x+ex)
33. y(x)=cos 3x−215sin 3x+15sin 2x
34. y(x)=cos x− sin x+12x sin x
35. y(x)=ex(2 cos x−52sin x)+12x+1
36. y(x)=1192(234+240x−9e−2x−33e2x−12x2−4x4)
37. y(x)=4−4ex+3xex+x−12x2ex+16x3ex
38. y(x)=185[e−x(176 cos x+197 sin x)−(6 cos 3x+7 sin 3x)]
39. y(x)=−3+3x−12x2+16x3+4e−x+xe−x
40. y(x)=14(5e−x+5ex+10 cos x−20)
41. yp(x)=255−450x+30x2+20x3+10x4−4x5
42. y(x)=10e−x+35e2x+210 cos x+390 sin x+yp(x) where yp(x) is the particular solution of Problem 41 .
43. (b) y(x)=c1 cos 2x+c2 sin 2x+14cos x−120cos 3x
44. y(x)=e−x/2[c1 cos (12x3)+c2 sin (12x3)]+126(−3 cos 2x+2 sin 2x)+1482(15 cos 4x+4 sin 4x)
45. y(x)=c1 cos 3x+c2 sin 3x+124−110cos 2x−156cos 4x
46. y(x)=c1 cos x+c2 sin x+116(3x cos x+3x2 sin x)+1128(3 sin 3x−4x cos 3x)
47. yp(x)=23ex
48. yp(x)=−112(6x+1)e−2x
49. yp(x)=x2e2x
50. yp(x)=116(4x cosh 2x−sinh 2x)
51. yp(x)=−14(cos 2x cos x− sin 2x sin x)+120(cos 5x cos 2x+ sin 5x sin 2x)=−15cos 3x (!)
52. yp(x)=−16x cos 3x
53. yp(x)=23x sin 3x+29(cos 3x) ln |cos 3x|
54. yp(x)=−1−(cos x) ln | csc x− cot x|
55. yp(x)=−18(1−x sin 2x)
56. yp(x)=−19ex(3x+2)
58. yp(x)=x3(ln x−1)
59. yp(x)=14x4
60. yp(x)=−725x4/3
61. yp(x)=ln x
62. yp(x)=−x2+x ln |1+x1−x|+12(1+x2) ln |1−x2|

Section 5.6

1. x(t)=2 cos 2t−2 cos 3t
2. x(t)=32sin 2t− sin 3t
3. x(t)=138388 cos (10t−α)+5 cos (5t−β) with α=2π−tan−1(1/186)≈6.2778 and β=tan−1(4/3)≈0.9273.
4. x(t)=2106 cos (5t−α)+10 cos 4t with α=π−tan−1(9/5)≈2.0779
5. x(t)=(x0−C) cos ω0t+C cos ωt, where C=F0/(k−mω2)
7. xsp(t)=1013cos (3t−α) with α=π−tan−1(12/5)≈1.9656
8. xsp(t)=425cos (5t−α) with α=2π−tan−1(3/4)≈5.6397
9. xsp(t)=340001cos (10t−α) with α=π+tan−1(199/20)≈4.6122
10. xsp(t)=1079361 cos (10t−α) with α=π+tan−1(171/478)≈3.4851
11. xsp(t)=104cos (3t−α) with α=π−tan−1(3)≈1.8925 xtr(t)=542e−2t cos (t−β) with β=2π−tan−1(7)≈4.8543
12. xsp(t)=5329cos (3t−α) with α=π+tan−1(2/5)≈3.5221 xtr(t)=25629e−3t cos (2t−β) with β=tan−1(5/2)≈1.1903
13. xsp(t)=3001469cos (10t−α) with α=π−tan−1(10/37)≈2.8776 xtr(t)=21133141469 e−t cos (5t−β) with β=2π−tan−1(421/12895)≈6.2505
14. xsp(t)=485 cos (t−α) with α=tan−1(22)≈1.5254xtr(t)=3665 e−4t cos (3t−β) with β=π+tan−1(52/31)≈4.1748
15. C(ω)=2/4+ω4; there is no practical resonance frequency.
16. C(ω)=10/25+6ω2+ω4; there is no practical resonance frequency.
17. C(ω)=50/2025−54ω2+ω4; there is practical resonance at frequency ω=33.
18. C(ω)=100/422500−1200ω2+ω4; there is practical resonance at frequency ω=106.
19. ω=384 rad/sec (approximately 3.12 Hz)
20. ω≈44.27 rad/sec (approximately 422.75 rpm)
21. ω0=(g/L)+(k/m)
22. ω0=k/(m+I/a2)
23. (a) Natural frequency: 10 rad/s (approximately 0.50 Hz); (b) amplitude: approximately 10.625 in.

Section 6.1

1. λ1=2: v1=(1, 1); λ2=3: v2=(2, 1)
2. λ1=−1: v1=(1, 1); λ2=2: v2=(2, 1)
3. λ1=2: v2=(1, 1); λ2=5: v1=(2, 1)
4. λ1=1: v2=(1, 1); λ2=2: v1=(3, 2)
5. λ1=1: v2=(1, 1); λ2=4: v1=(3, 2)
6. λ1=2: v2=(1, 1); λ2=3: v1=(4, 3)
7. λ1=2: v2=(1, 1); λ2=4: v1=(4, 3)
8. λ1=−2: v2=(2, 3); λ2=−1: v1=(3, 4)
9. λ1=3: v1=(2, 1); λ2=4: v2=(5, 2)
10. λ1=4: v1=(2, 1); λ2=5: v2=(5, 2)
11. λ1=4: v2=(2, 3); λ2=5: v1=(5, 7)
12. λ1=3: v2=(3, 2); λ2=4: v1=(5, 3)
13. λ1=0: v1=(0,−1, 2); λ2=1: v2=(0,−1, 3); λ3=2: v3=(1, 0, 2)
14. λ1=0: v1=(0,−1, 2); λ2=2: v2=(0,−1, 3); λ3=5: v3=(1, 0, 2)
15. λ1=0: v1=(1, 1, 0); λ2=1: v2=(2, 1, 1); λ3=2: v3=(1, 0, 2)
16. λ1=0: v1=(1, 1, 1); λ2=1: v2=(1, 1, 0); λ3=3: v3=(−1, 0, 2)
17. λ1=1: v1=(1, 0, 1); λ2=2: v2=(−1, 1, 2); λ3=3: v3=(1, 0, 0)
18. λ1=1: v1=(1, 0, 3); λ2=2: v2=(0,−1, 3); λ3=3: v3=(0,−2, 5)
19. λ1=λ2=1: v1=(1, 0, 1), v2=(−3, 1, 0); λ3=3, v3=(1, 0, 0)
20. λ1=λ2=1: v1=(1, 0, 2), v2=(3, 2, 0); λ3=2, v3=(0,−2, 5)
21. λ1=1: v1=(1, 1, 0); λ2=λ3=2: v2=(3, 2, 0), v3=(−1, 0, 2)
22. λ1=λ2=−1: v1=(1, 1, 0), v2=(−1, 0, 2); λ3=2, v3=(1, 1, 1)
23. λ1=1: v1=(1, 0, 0, 0); λ2=2: v2=(2, 1, 0, 0); λ3=3: v3=(3, 2, 1, 0); λ4=4: v4=(4, 3, 2, 1)
24. λ1=λ2=1: v1=(1, 0, 0, 0), v2=(0, 1, 0, 0); λ3=λ4=3: v3=(0, 0, 0, 1), v4=(2, 2, 1, 0)
25. λ1=λ2=1: v1=(1, 0, 0, 0), v2=(0, 1, 0, 0); λ3=λ4=2: v3=(0, 0, 0, 1), v4=(1, 1, 1, 0)
26. λ1=−2: v1=(1, 0, 0, 2); λ2=−1: v2=(0, 0, 1, 0); λ3=1: v3=(1, 0, 0, 1); λ4=2: v4=(0, 1, 0, 0)
27. λ1=−i: v1=(+i, 1); λ2=+i: v2=(−i, 1)
28. λ1=−6i: v1=(−i, 1); λ2=+6i: v2=(+i, 1)
29. λ1=−6i: v1=(−i, 2); λ2=+6i: v2=(+i, 2)
30. λ1=−12i: v1=(−i, 1); λ2=+12i: v2=(+i, 1)
31. λ1=−12i: v1=(+2i, 1); λ2=+12i: v2=(−2i, 1)
32. λ1=−12i: v1=(−i, 3); λ2=+12i: v2=(+i, 3)
40. We find that Tr A=12 and detA=60, so the characteristic polynomial of the given matrix A is p(λ)=−λ3+12λ2+c1λ+60. Eigenvalues and eigenvectors: λ1=3: v1=(3, 2, 1); λ2=4: v2=(5, 7, 7); λ3=5: v3=(−1, 1, 2)
41. We find that Tr A=8 and detA=−60, so the characteristic polynomial of the given matrix A is p(λ)=λ4−8λ3+c2λ2+c1λ−60. Eigenvalues and eigenvectors: λ1=−2: v1=(1, 0, 1, 2); λ2=2: v2=(3, 4, 0, 3); λ3=3: v3=(3, 1, 2, 4); λ4=5: v4=(1, 1, 0, 1)

Section 6.2

1. λ1=1, λ2=3; P=[1211], D=[1003]
2. λ1=0, λ2=2; P=[1312], D=[0002]
3. λ1=2, λ2=3; P=[1312], D=[2003]
4. λ1=1, λ2=2; P=[1413], D=[1002]
5. λ1=1, λ2=3; P=[1413], D=[1003]
6. λ1=1, λ2=2; P=[2334], D=[1002]
7. λ1=1, λ2=2; P=[2512], D=[1002]
8. λ1=1, λ2=2; P=[3523], D=[1002]
9. The double eigenvalue λ1=λ2=1 has only the single associated eigenvector v1=(2, 1), so the matrix A is not diagonalizable.
10. The double eigenvalue λ1=λ2=2 has only the single associated eigenvector v1=(1, 1), so the matrix A is not diagonalizable.
11. The double eigenvalue λ1=λ2=2 has only the single associated eigenvector v1=(−1, 3), so the matrix A is not diagonalizable.
12. The double eigenvalue λ1=λ2=−1 has only the single associated eigenvector v1=(−3, 4), so the matrix A is not diagonalizable.
13. λ1=1, λ2=λ3=2; P=[103001010], D=[100020002]
14. λ1=λ2=0, λ3=1;P=[−111011210], D=[000000001]
15. λ1=0, λ2=λ3=1; P=[1−13102020], D=[000010001]
16. λ1= λ2=1, λ3=3; P=[01−1010102], D=[100010003]
17. λ1=−1, λ2=1, λ3=2; P=[1−11101021], D=[−100010002]
18. λ1=1, λ2=2, λ3=3; P=[1−11101021], D=[100020003]
19. λ1=1, λ2=2, λ3=3; P=[11−1110102], D=[100020003]
20. λ1=2, λ2=5, λ3=6; P=[1000−1−2335], D=[200050006]
21. The triple eigenvalue λ1=λ2=λ3=1 has only the two associated eigenvectors v1=(0, 0, 1) and v2=(1, 1, 0), so the matrix A is not diagonalizable.
22. The triple eigenvalue λ1=λ2=λ3=1 has only the single associated eigenvector v1=(1, 1, 1), so the matrix A is not diagonalizable.
23. The eigenvalues λ1=λ2=1 and λ3=2 have only the two associated eigenvectors v1=(1, 1, 1) and v3=(1, 1, 0), so the matrix A is not diagonalizable.
24. The eigenvalues λ1=1 and λ2=λ3=2 have only the two associated eigenvectors v1=(1, 1, 0) and v2=(1, 1, 1), so the matrix A is not diagonalizable.
25. λ1=λ2=−1, λ3=λ4=1; P=[0101011001001000], D=[−10000−10000100001]
26. λ1=λ2=λ3=1, λ4=2; P=[0011010000010001], D=[1000010000100002]
27. The eigenvalues λ1=λ2=λ3=1 and λ4=2 have only the two associated eigenvectors v1=(1, 0, 0, 0) and v4=(1, 1, 1, 1), so the matrix A is not diagonalizable.
28. The eigenvalues λ1=λ2=1 and λ3=λ4=2 have only the two associated eigenvectors v1=(1, 0, 0, 0) and v3=(1, 1, 1, 0), so the matrix A is not diagonalizable.

Section 6.3

1. P=[1211], D=[1002], A5=[63−6231−30]
2. P=[1211], D=[−1002], A5=[65−6633−34]
3. P=[1312], D=[0002], A5=[96−9664−64]
4. P=[1312], D=[1002], A5=[94−9362−61]
5. P=[1413], D=[1002], A5=[125−12493−92]
6. P=[2511], D=[1002], A5=[156−31062−123]
7. P=[103001010], D=[100020002], A5=[193003200032]
8. P=[011100201], D=[100010002], A5=[1−62310100−6232]
9. P=[11−3001010], D=[100020002], A5=[1−933103200032]
10. P=[1−13102020], D=[100020002], A5=[94−933162−61310032]
11. P=[00−1−1−2−423587], D=[−100000001], A10=[10078−5−2−195156]
12. P=[113205050], D=[−100010001], A10=[100010001]
13. P=[−111011201], D=[−100000001], A10=[3−312−21001]
14. P=[112011201], D=[−100000001], A10=[3−312−21001]
15. A2−3A+2I=0, A3=[29−2821−20], A4=[61−6045−44], A−1=12[−24−35]
16. A2−3A+2I=0, A3=[36−7014−27], A4=[76−15030−59], A−1=12[−310−26]
17. −A3+5A2−8A+4I=0, A3=[1210080008], A4=[145001600016], A−1=12[2−30010001]
18. −A3+4A2−5A+2I=0, A3=[1−1470100−148], A4=[1−30150100−3016], A−1=12[22−1020021]
19. −A3+5A2−8A+4I=0, A3=[1−217080008], A4=[1−451501600016], A−1=12[23−1010001]
20. −A3+5A2−8A+4I=0, A3=[22−21714−137008], A4=[46−451530−29150016], A−1=12[−13−1−24−1001]
21. −A3+A=0, A3=A=A=[10065221−15−6], A4=A2=[10078−5−2−195156]. Because λ=0 is an eigenvalue, A is singular and A−1 does not exist.
22. −A3+A2+A−I=0, A3=A=[11−6−220−11−4001]=A, A4=[100010001]=I, A−1=A
23. −A3+A=0, A3=A=[1−112−214−41], A4=A2=[3−312−21001]. Because λ=0 is an eigenvalue, A is singular and A−1 does not exist.
24. −A3+A=0, A3=A= [5−5−32−2−14−4−3], A4=A2=[3−3−12−2−1001]. Because λ=0 is an eigenvalue, A is singular and A−1 does not exist.
25. xk=Akx0=[1−111][1004/5]k12[11−11]x0→=(C0+S0)[1/21/2] as k→∞. The long-term distribution of population is 50% city, 50% suburban.
26. xk=Akx0=[1−131][1004/5]k14[11−31]x0→=(C0+S0)[1/43/4] as k→∞. The long-term distribution of population is 25% city, 75% suburban.
27. xk=Akx0=[3−151][1003/5]k18[11−53]x0→=(C0+S0)[3/85/8] as k→∞. The long-term distribution of population is 3/8 city, 5/8 suburban.
28. xk=Akx0=[1−121][1007/10]k13[11−21]x0→=(C0+S0)[1/32/3] as k→∞. The long-term distribution of population is 1/3 city, 2/3 suburban.
29. xk=Akx0=[1−121][10017/20]k13[11−21]x0→=(C0+S0)[1/32/3] as k→∞. The long-term distribution of population is 1/3 city, 2/3 suburban.
30. xk=Akx0=[3−141][10013/20]k17[11−43]x0→=(C0+S0)[3/74/7] as k→∞. The long-term distribution of population is 3/7 city, 4/7 suburban.
31. xk=Akx0=[5542][1004/5]k110[−254−5]x0→=[2.5R0−F02R0−0.8F0] as k→∞. The fox-rabbit population approaches a stable situation with 2.5R0−F0 foxes and 2R0−0.8F0 rabbits.
32. xk=Akx0=[10271][19/200017/20]k14[−127−10]x0→=[00] as k→∞. The fox and rabbit population both die out.
33. xk=Akx0=[101093][21/20003/4]k160[−3109−10]x0≈160(1.05)k(10R0−3F0)[109] as when k is sufficiently large. The fox and rabbit populations are both increasing at 5% per year, with 10 foxes for each 9 rabbits.
34. A=PDP−1=[41−3056−41]. If n is even, then Dn=I so An=PDnP−1=PIP−1=I. If n is odd, then An=An−1A=IA=A. Thus A99=A and A100=I.
35. λ=±1 implies that Dn=I if n is even, in which case An=PDnP−1=I.
36. A2=I, so A3=A2A=IA=A, A4=A3A=A2=I, and so forth.
37. A2=−I, so A3=A2A=−IA=−A, A4=A3A=−A2=I, and so forth.
38. If B=[0100] so B2=0, and it follows that An=I+nB=[1n01].

Section 7.1

1. x1′=x2, x2′=−7x1−3x2+t2
2. x1′=x2, x2′=x3, x3′=x4,x4′=−x1+3x2−6x3+ cos 3t
3. x1′=x2, t2x2′=(1−t2)x1−tx2
4. x1′=x2, x2′=x3, t3x3′=−5x1−3tx2+2t2x3+ ln t
5. x1′=x2, x2′=x3, x3′=x22+ cos x1
6. x1′=x2, x2′=5x1−4y1, y1′=y2, y2′=−4x1+5y1
7. x1′=x2, y1′=y2, x2′=−kx1·(x12+y12)−3/2, y2′=−ky1·(x12+y12)−3/2
8. x1′=x2, x2′=−4x1+2y1−3x2 y1′=y2, y2′=3x1−y1−2y2+ cos t
9. x1′=x2, y1′=y2, z1′=z2, x2′=3x1−y1+2z1, y2′=x1+y1−4z1, z2′=5x1−y1−z1
10. x1′=x2, x2′=x1(1−y1) y1′=y2, y2′=y1(1−x1)
11. x(t)=A cos t+B sin t, y(t)=B cos t−A sin t
12. x(t)=Aet+Be−t, y(t)=Aet−Be−t
13. x(t)=A cos 2t+B sin 2t, y(t)=−B cos 2t+A sin 2t; x(t)=cos 2t, y(t)=sin 2t
14. x(t)=A cos 10t+B sin 10t, y(t)=B cos 10t−A sin 10t; x(t)=3 cos 10t+4 sin 10t, y(t)=4 cos 10t−3 sin 10t
15. x(t)=A cos 2t+B sin 2t, y(t)=4B cos 2t−4A sin 2t
16. x(t)=A cos 4t+B sin 4t, y(t)=12B cos 4t−12A sin 4t
17. x(t)=Ae−3t+Be2t, y(t)=−3Ae−3t+2Be2t; x(t)=e2t, y(t)=2e2t
18. x(t)=Ae−2t+Be−5t, y(t)=2Ae−2t+5Be−5t; A=173 and B=−113 in the particular solution.
19. x(t)=−e−2t sin 3t, y(t)=e−2t(3 cos 3t+2 sin 3t)
20. x(t)=(A+Bt)e3t, y(t)=(3A+B+3Bt)e3t

Section 7.2

1. (AB)′=[1−8t+18t21+2t−12t2+32t33+3t2−4t38t+3t2+4t3]
3. x=[xy], P(t)=[0−330], f(t)=[00]
5. x=[xy], P(t)=[245−1], f(t)=[3et−t2]
7. x=[xyz], P(t)=[011101110], f(t)=[000]
9. x=[xyz], P(t)=[3−4110−306−7], f(t)=[tt2t3]
11. x=[x1x2x3x4], P(t)=[0100002000034000], f(t)=[0000]
13. W(t)=e3t; x(t)=[2c1et+c2e2t−3c1et−c2e2t]
15. W(t)≡4; x(t)=[c1e2t+c2e−2tc1e2t+5c2e−2t]
17. W(t)=7e−3t; x(t)=[3c1e2t+c2e−2t2c1e2t+3c2e−2t]
19. W(t)≡3; x(t)=[c1e2t+c2e−tc1e2t+c3e−tc1e2t−(c2+c3)e−t]
21. W(t)=e2t; x(t)=[3c1e−2t+c2et+c3et−2c1e−2t−c2et−c3et2c1e−2t+c2et]
23. x=2x1−x2
25. x=15x1−4x2
27. x=x1+2x2+x3
29. x=3x1−3x2−5x3
31. x=3x1+7x2+x3−2x4
32. x=13x1+41x2+3x3−12x4
33. (a) x2=tx1, so neither is a constant multiple of the other. (b) W(x1,x2)≡0, whereas Theorem 2 implies that W≠0 if x1 and x2 were independent solutions of a system of the indicated form.

Section 7.3

1. x1(t)=c1e−t+c2e3t, x2(t)=−c1e−t+c2e3t
2. x1(t)=c1e−t+3c2e4t, x2(t)=−c1e−t+2c2e4t
3. General solution x1(t)=c1e−t+4c2e6t, x2(t)=−c1e−t+3c2e6t Particular solution x1(t)=17(−e−t+8e6t), x2(t)=17(e−t+6e6t).
4. x1(t)=c1e−2t+c2e5t, x2(t)=−6c1e−2t+c2e5t
5. x1(t)=c1e−t+7c2e5t, x2(t)=c1e−t+c2e5t
6. General solution x1(t)=5c1e3t+c2e4t, x2(t)=−6c1e3t−c2e4t Particular solution x1(t)=−5e3t+6e4t, x2(t)=6e3t−6e4t.
7. x1(t)=c1et+2c2e−9t, x2(t)=c1et−3c2e−9t
8. x1(t)=5c1 cos 2t+5c2 sin 2t, x2(t)=(c1−2c2) cos 2t+(2c1+c2) sin 2t
9. General solution x1(t)=5c1 cos 4t+5c2 sin 4t, x2(t)=c1(2 cos 4t+4 sin 4t)+c2(2 sin 4t−4 cos 4t). Particular solution x1(t)=2 cos 4t−114sin 4t, x2(t)=3 cos 4t+12sin 4t
10. x1(t)=−2c1 cos 3t−2c2 sin 3t, x2(t)=(3c1+3c2) cos 3t+(3c2−3c1) sin 3t
11. General solution x1(t)=et(c1 cos 2t−c2 sin 2t), x2(t)=et(c1 sin 2t+c2 cos 2t) Particular solution x1(t)=−4et sin 2t, x2(t)=4et cos 2t
12. x1(t)=e2t(−5c1 cos 2t−5c2 sin 2t), x2(t)=e2t[(c1+2c2) cos 2t+(−2c1+c2) sin 2t]
13. x1(t)=3e2t(c1 cos 3t−c2 sin 3t), x2(t)=e2t[(c1+c2) cos 3t+(c1−c2) sin 3t]
14. x1(t)=e3t(c1 cos 4t+c2 sin 4t), x2(t)=e3t(c1 sin 4t−c2 cos 4t)
15. x1(t)=5e5t(c1 cos 4t−c2 sin 4t), x2(t)=e5t[(2c1+4c2) cos 4t+(4c1−2c2) sin 4t]
16. x1(t)=c1e−10t+2c2e−100t, x2(t)=2c1e−10t−5c2e−100t
17. x1(t)=c1e9t+c2e6t+c3, x2(t)=c1e9t−2c2e6t, x3(t)=c1e9t+c2e6t−c3
18. x1(t)=c1e9t+4c3, x2(t)=2c1e9t+c2e6t−c3, x3(t)=2c1e9t−c2e6t−c3
19. x1(t)=c1e6t+c2e3t+c3e3t, x2(t)=c1e6t−2c2e3t, x3(t)=c1e6t+c2e3t−c3e3t
20. x1(t)=c1e9t+c2e6t+c3e2t, x2(t)=c1e9t−2c2e6t, x3(t)=c1e9t+c2e6t−c3e2t
21. x1(t)=6c1+3c2et+2c3e−t, x2(t)=2c1+c2et+c3e−t, x3(t)=5c1+2c2et+2c3e−t
22. x1(t)=c2et+c3e3t, x2(t)=c1e−2t−c2et−c3e3t, x3(t)=−c1e−2t+c3e3t
23. x1(t)=c1e2t+c3e3t, x2(t)=−c1e2t+c2e−2t−c3e3t, x3(t)=−c2e−2t+c3e3t
24. x1(t)=c1et+c2(2 cos 2t− sin 2t)+c3(cos 2t+2 sin 2t) x2(t)=−c1et−c2(3 cos 2t+ sin 2t)+c3(cos 2t−3 sin 2t) x3(t)=c2(3 cos 2t+ sin 2t)+c3(3 sin 2t− cos 2t)
25. x1(t)=c1+e2t[(c2+c3) cos 3t+(−c2+c3) sin 3t], x2(t)=−c1+2e2t(−c2 cos 3t−c3 sin 3t), x3(t)=2e2t(c2 cos 3t+c3 sin 3t)
26. x1(t)=4e3t−e−t(4 cos t−sin t), x2(t)=9e3t−e−t(9 cos t+2 sin t), x3(t)=17e−t cos t
27. x1(t)=15e−0.2t, x2(t)=15(e−0.2t−e−0.4t). The maximum amount ever in tank 2 is x2(5 ln 2)=3.75 lb.
28. x1(t)=15e−0.4t, x2(t)=40(−e−0.4t+e−0.25t). The maximum amount ever in tank 2 is about 6.85 lb.
29. x1(t)=10+5e−0.6t, x2(t)=5−5e−0.6t
30. x1(t)=15c1+c2e−0.65t,x2(t)=8c1−c2e−0.65t.
31. x1(t)=27e−t, x2(t)=27e−t−27e−2t, x3(t)=27e−t−54e−2t+27e−3t.

The maximum amount of salt everin tank 3 is x3(ln 3)=4 pounds.
32. x1(t)=45e−3t, x2(t)=−135e−3t+135e−2t, x3(t)=135e−3t−270e−2t+135e−t.

The maximum amount of salt ever in tank 3 is x3(ln 3)=20 pounds.
33. x1(t)=45e−4t, x2(t)=90e−4t−90e−6t, x3(t)=−270e−4t+135e−6t+135e−2t.

The maximum amount of saltever in tank 3 is x3(12 ln 3)=20 pounds.
34. x1(t)=40e−3t, x2(t)=60e−3t−60e−5t, x3(t)=−150e−3t+75e−5t+75e−t.

The maximum amount of salt ever in tank 3 is x3(12 ln 5)≈21.4663 pounds.
35. x1(t)=10−17(55e−18t−216e−11t), x2(t)=3−17(165e−18t−144e−11t), x3(t)=20+17(220e−18t−360e−11t). The limiting amounts of salt in tanks 1, 2, and 3 are 10 lb, 3 lb, and 20 lb.
36. x1(t)=4+e−3t/5[14 cos (3t/10)−2 sin (3t/10)], x2(t)=10−e−3t/5[10 cos (3t/10)−10 sin (3t/10)], x3(t)=4−e−3t/5[4 cos (3t/10)+8 sin (3t/10)]. The limiting amounts of salt in tanks 1, 2, and 3 are 4 lb, 10 lb, and 4 lb.
37. x1(t)=30+e−3t[25 cos (t2)+102 sin (t2)], x2(t)=10−e−3t[10 cos (t2)−2522 sin (t2)], x3(t)=15−e−3t[15 cos (t2)+4522 sin (t2)]. The limiting amounts of salt in tanks 1, 2, and 3 are 30 lb, 10 lb, and 15 lb.
38. x1(t)=c1et, x2(t)=−2c1et+c2e2t, x3(t)=3c1et−3c2e2t+c3e3t, x4(t)=−4c1et+6c2e2t−4c3e3t+c4e4t
39. x1(t)=3c1et+c4e−2t, x2(t)=−2c1et+c3e2t−c4e−2t, x3(t)=4c1et+c2e−t, x4(t)=c1et
40. x1(t)=c1e2t, x2(t)=−3c1e2t+3c2e−2t−c4e−5t, x3(t)=c3e5t, x4(t)=−c2e−2t−3c3e5t
41. x1(t)=2e10t+e15t=x4(t), x2(t)=−e10t+2e15t=x3(t)
42. x(t)=c1[3−12]+c2[111]e2t+c3[2−31]e5t
43. x(t)=c1[3−15]e−2t+c2[111]e4t+c3[1−13]e8t
44. x(t)=c1[3−22]e−3t+c2[715]e6t+c3[5−33]e12t
45. x(t)=c1[111−1]e−3t+c2[12−11]+c3[2111]e3t+c4[1−12−1]e6t
46. x(t)=c1[32−11]e−4t+c2[122−1]e2t+c3[11−11]e4t+c4[3−23−3]e8t
47. x(t)=c1[221−1]e−3t+c2[12−11]e3t+c3[2111]e6t+c4[1−12−1]e9t
48. x(t)=c1[12−12]e16t+c2[251−1]e32t+c3[3−112]e48t+c4[112−3]e64t
49. x(t)=c1[10311]e−3t+c2[030−11]+c3[17111]e3t+c4[01011]e6t+c5[20521]e9t
50. x(t)=c1[011101]e−7t+c2[100011]e−4t+c3[010101]e3t+c4[001010]e5t+c5[110001]e9t+[001−1−10]e11t

Section 7.4

Note that phase portraits for Problems 1-16 are found in the answers for Section 7.2.

1. Saddle point (real eigenvalues of opposite sign)
2. Saddle point (real eigenvalues of opposite sign)
3. Saddle point (real eigenvalues of opposite sign)
4. Saddle point (real eigenvalues of opposite sign)
5. Saddle point (real eigenvalues of opposite sign)
6. Improper nodal source (distinct positive real eigenvalues)
7. Saddle point (real eigenvalues of opposite sign)
8. Center (pure imaginary eigenvalues)
9. Center (pure imaginary eigenvalues)
10. Center (pure imaginary eigenvalues)
11. Spiral source (complex conjugate eigenvalues with positive real part)
12. Spiral source (complex conjugate eigenvalues with positive real part)
13. Spiral source (complex conjugate eigenvalues with positive real part)
14. Spiral source (complex conjugate eigenvalues with positive real part)
15. Spiral source (complex conjugate eigenvalues with positive real part)
16. Improper nodal sink (distinct negative real eigenvalues)
17. Center; pure imaginary eigenvalues
18. Improper nodal source; distinct positive real eigenvalues; v1=[01]T, v2=[−11]T
19. Saddle point; real eigenvalues of opposite sign; v1=[01]T corresponds to the negativeeigen value and v2=[−11]T to the positive one.
20. Spiral source; complex conjugate eigenvalues with positive real part
21. Proper nodal source; repeated positive real eigenvalue with linearly independent eigenvectors
22. Parallel lines; one zero and one negative real eigenvalue
23. Spiral sink; complex conjugate eigenvalues with negative real part
24. Improper nodal sink; distinct negative real eigenvalues; v1=[11]T,v2=[−14]T
25. Saddle point; real eigenvalues of opposite sign; v1=[11]T corresponds to the positive eigenvalue and v2=[4−1]T to the negative one.
26. Center; pure imaginary eigenvalues
27. Improper nodal source; distinct positive real eigenvalues; v1=[23]T, v2=[2−1]T.
28. Spiral sink; complex conjugate eigenvalues with negative real part

Section 7.5

1. The natural frequencies are ω0=0 and ω1=2. In the degenerate natural mode with “frequency” ω0=0 the two masses move linearly with x1(t)=x2(t)=a0+b0t. At frequency ω1=2 they oscillate in opposite directions with equal amplitudes.
2. The natural frequencies are ω1=1 and ω2=3. In the natural mode with frequency ω1, the two masses m1 and m2 move in the same direction with equal amplitudes of oscillation. At frequency ω2 they move in opposite directions with equal amplitudes.
3. The natural frequencies are ω1=1 and ω2=2. In the natural mode with frequency ω1, the two masses m1 and m2 move in the same direction with equal amplitudes of oscillation. In the natural mode with frequency ω2 they move in opposite directions with the amplitude of oscillation of m1 twice that of m2.
4. The natural frequencies are ω1=1 and ω2=5. In the natural mode with frequency ω1, the two masses m1 and m2 move in the same direction with equal amplitudes of oscillation. At frequency ω2 they move in opposite directions with equal amplitudes.
5. The natural frequencies are ω1=2 and ω2=2. In the natural mode with frequency ω1, the two masses m1 and m2 move in the same direction with equal amplitudes of oscillation. At frequency ω2 they move in opposite directions with equal amplitudes.
6. The natural frequencies are ω1=2 and ω2=8. In the natural mode with frequency ω1, the two masses m1 and m2 move in the same direction with equal amplitudes of oscillation. In the natural mode with frequency ω2 they move in opposite directions with the amplitude of oscillation of m1 twice that of m2.
7. The natural frequencies are ω1=2 and ω2=4. In the natural mode with frequency ω1, the two masses m1 and m2 move in the same direction with equal amplitudes of oscillation. At frequency ω2 they move in opposite directions with equal amplitudes.
8. x1(t)=2 cos t+3 cos 3t−5 cos 5t, x2(t)=2 cos t−3 cos 3t+ cos 5t. We have a superposition of three oscillations, in which the two masses move (1) in the same direction with frequency ω1=1 and equal amplitudes; (2) in opposite directions with frequency ω2=3 and equal amplitudes; (3) in opposite directions with frequency ω3=5 and with the amplitude of motion of m1 being 5 times that of m2.
9. x1(t)=5 cos t−8 cos 2t+3 cos 3t, x2(t)=5 cos t+4 cos 2t−9 cos 3t. We have a superposition of three oscillations, in which the two masses move (1) in the same direction with frequency ω1=1 and equal amplitudes; (2) in opposite directions with frequency ω2=2 and with the amplitude of motion of m1 being twice that of m2; (3) in opposite directions with frequency ω3=3 and with the amplitude of motion of m2 being 3 times that of m1.
10. x1(t)=−15 cos 2t+ cos 4t+14 cos t, x2(t)=−15 cos 2t− cos 4t+16 cos t. We have a superposition of three oscillations, in which the two masses move (1) in the same direction with frequency ω1=1 and with the amplitude of motion of m2 being 8/7 times that of m1; (2) in the same direction with frequency ω2=2 and equal amplitudes; (3) in opposite directions with frequency ω3=4 and equal amplitudes.
11. (a) The natural frequencies are ω1=6 and ω2=8. In mode 1 the two masses oscillate in the same direction with frequency ω1=6 and with the amplitude of motion of m1 being twice that of m2. In mode 2 the two masses oscillate in opposite directions with frequency ω2=8 and with the amplitude of motion of m2 being 3 times that of m1. (b) x(t)=2 sin 6t+19 cos 7t, y(t)=sin 6t+3 cos 7t We have a superposition of (only two) oscillations, in which the two masses move (1) in the same direction with frequency ω1=6 and with the amplitude of motion of m1 being twice that of m2; (2) in the same direction with frequency ω3=7 and with the amplitude of motion of m1 being 19/3 times that of m2.
12. The system’s three natural modes of oscillation have (1) natural frequency ω1=2 with amplitude ratios 1:0:−1; (2) natural frequency ω2=2+2 with amplitude ratios 1 : −2 : 1; (3) natural frequency ω3=2−2 with amplitude ratios 1 : 2 : 1.
13. The system’s three natural modes of oscillation have (1) natural frequency ω1=2 with amplitude ratios 1 : 0 : −1; (2) natural frequency ω2=4+22 with amplitude ratios 1 : −2 : 1; (3) natural frequency ω3=4−22 with amplitude ratios 1 : 2 : 1.
15. x1(t)=23cos 5t−2 cos 53 t+43cos 10t, x2(t)=43cos 5t+4 cos 53 t+163cos 10t.

We have a superposition of two oscillations with the natural frequencies ω1=5 and ω2=53 and a forced oscillation with frequency ω=10. In each of the two natural oscillations the amplitude of motion of m2 is twice that of m1, while in the forced oscillation the amplitude of motion of m2 is four times that of m1.
20. x1′(t)=−v0, x2′(t)=0, x1′(t)=v0 for t>π/2
21. x1′(t)=−v0, x2′(t)=0, x1′(t)=2v0 for t>π/2
22. x1′(t)=−2v0, x2′(t)=v0, x1′(t)=v0 for t>π/2
23. x1′(t)=2v0, x2′(t)=2v0, x1′(t)=3v0 for t>π/2
24. (a) ω1≈1.0293 Hz; ω2≈1.7971 Hz. (b) v1≈28 mi/h; v2≈49 mi/h
27. ω1=210, v1≈40.26 (ft/s (about 27 mi/h), ω2=55, v2≈71.18 ft/s (about 49 mi/h)
28. ω1≈6.1311, v1≈39.03 ft/s (about 27 mi/h) ω2≈10.3155, v2≈65.67 ft/s (about 45 mi/h)
29. ω1≈5.0424, v1≈32.10 ft/s (about 22 mi/h), ω2≈9.9158, v2≈63.13 ft/s (about 43 mi/h)

Section 7.6

1. Repeated eigenvalue λ=−3, eigenvector v=[1−1]T; x1(t)=(c1+c2+c2t)e−3t, x2(t)=(−c1−c2t)e−3t
2. Repeated eigenvalue λ=2, single eigenvector v=[11]T; x1(t)=(c1+c2+c2t)e2t, x2(t)=(c1+c2t)e2t
3. Repeated eigenvalue λ=3, eigenvector v=[−22]T; x1(t)=(−2c1+c2−2c2t)e3t, x2(t)=(2c1+2c2t)e3t
4. Repeated eigenvalue λ=4, single eigenvector v=[−11]T; x1(t)=(−c1+c2−c2t)e4t, x2(t)=(c1+c2t)e4t
5. Repeated eigenvalue λ=5, eigenvector v=[2−4]T; x1(t)=(2c1+c2+2c2t)e5t, x2(t)=(−4c1−4c2t)e5t
6. Repeated eigenvalue λ=5, single eigenvector v=[−44]T; x1(t)=(−4c1+c2−4c2t)e5t, x2(t)=(4c1+4c2t)e5t
7. Eigenvalues λ=2, 2, 9 with three linearly independent eigenvectors; x1(t)=c1e2t+c2e2t, x2(t)=c1e2t+c3e9t, x3(t)=c2e2t
8. Eigenvalues λ=7, 13, 13 with three linearly independent eigenvectors; x1(t)=2c1e7t−c3e13t, x2(t)=−3c1e7t+c3e13t, x3(t)=c1e7t+c2e13t
9. Eigenvalues λ=5, 5, 9 with three linearly independent eigenvectors; x1(t)=c1e5t+7c2e5t+3c3e9t, x2(t)=2c1e5t, x3(t)=2c2e5t+c3e9t
10. Eigenvalues λ=3, 3, 7 with three linearly independent eigenvectors; x1(t)=5c1e3t−3c2e3t+2c3e7t, x2(t)=2c1e3t+c3e7t, x3(t)=c2e3t
11. Triple eigenvalue λ=−1 of defect 2; x1(t)=(−2c2+c3−2c3t)e−t, x2(t)=(c1−c2+c2t−c3t+12c3t2)e−t, x3(t)=(c2+c3t)e−t
12. Triple eigenvalue λ=−1 of defect 2; x1(t)=e−t(c1+c3+c2t+12c3t2) x2(t)=e−t(c1+c2t+12c3t2), x3(t)=e−t(c2+c3t)
13. Triple eigenvalue λ=−1 of defect 2; x1(t)=(c1+c2t+12c3t2)e−t, x2(t)=(2c2+c3+2c3t)e−t, x3(t)=(c2+c3t)e−t
14. Triple eigenvalue λ=−1 of defect 2; x1(t)=e−t(5c1+c2+c3+5c2t+c3t+52c3t2), x2(t)=e−t(−25c1−5c2−25c2t−5c3t−252c3t2), x3(t)=e−t(−5c1+4c2−5c2t+4c3t−52c3t2)
15. Triple eigenvalue λ=1 of defect 1; x1(t)=(3c1+c3−3c3t)et, x2(t)=(−c1+c3t)et, x3(t)=(c2+c3t)et
16. Triple eigenvalue λ=1 of defect 1; x1(t)=et(3c1+3c2+c3) x2(t)=et(−2c1−2c3t), x3(t)=et(−2c2+2c3t)
17. Triple eigenvalue λ=1 of defect 1; x1(t)=(2c1+c2)et, x2(t)=(−3c2+c3+6c3t)et, x3(t)=−9(c1+c3t)et
18. Triple eigenvalue λ=1 of defect 1; x1(t)=et(−c1−2c2+c3), x2(t)=et(c2+c3t), x3(t)=et(c1−2c3t)
19. Double eigenvalues λ=−1 and λ=1, with four linearly independent solutions; x1(t)=c1e−t+c4et, x2(t)=c3et, x3(t)=c2e−t+3c4et, x4(t)=c1e−t−2c3et
20. Eigenvalue λ=2 with multiplicity 4 and defect 3; x1(t)=(c1+c3+c2t+c4t+12c3t2+16c4t3)e2t, x2(t)=(c2+c3t+12c4t2)e2t, x3(t)=(c3+c4t)e2t, x4(t)=c4e2t
21. Eigenvalue λ=1 with multiplicity 4 and defect 2; x1(t)=(−2c2+c3−2c3t)et, x2(t)=(c2+c3t)et, x3(t)=(c2+c4+c3t)et, x4(t)=(c1+c2t+12c3t2)et
22. Eigenvalue λ=1 with multiplicity 4 and defect 2; x1(t)=(c1+3c2+c4+c2t+3c3t+12c3t2)et, x2(t)=−(2c2−c3+2c3t)et, x3(t)=(c2+c3t)et, x4(t)=−(2c1+6c2+2c2t+6c3t+c3t2)et
23. x(t)=c1v1e−t+(c2v2+c3v3)e3t with v1=[1−12]T, v2=[409]T, v3=[021]T
24. x(t)=c1v1e−t+(c2v2+c3v3)e3t with v1=[53−3]T, v2=[40−1]T, v3=[2−10]T
25. x(t)=[c1v1+c2(v1t+v2)+c3(12v1t2+v2t+v3)]e2t with v1=[−10−1]T, v2=[−4−10]T, and v3=[100]T
26. x(t)=[c1v1+c2(v1t+v2)+c3(12v1t2+v2t+v3)]e3t with v1=[022]T, v3=[21−3]T, and v3=[100]T
27. x(t)=[c1v1+c2(v1t+v2)+c3v3]e2t with v1=[−538]T, v2=[100]T, v3=[100]T
28. x(t)=[c1v1+c2(v1t+v2)+c3(12v1t2+v2t+v3)]e2t with v1=[119−2890]T, v2=[−173417]T, and v3=[100]T
29. x(t)=[c1v1+c2(v1t+v2)]e−t+[c3v3+c4(v3t+v4)]e2t with v1=[1−3−1−2]T, v2=[0100]T, v3=[0−110]T, v4=[0021]T
30. x(t)=[c1v1+c2(v1t+v2)]e−t+[c3v3+c4(v3t+v4)]e2t, with v1=[01−1−3]T, v2=[0012]T, v3=[−1000]T, v4=[0035]T
31. x(t)=[c1v1+c2(v1t+v2)+c3(12v1t2+v2t+v3)+c4v4]et with v1=[427−21−42]T, v2=[3422−10−27]T, v3=[1000]T, v4=[0130]T
32. x(t)=(c1v1+c2v2)e2t+(c3v3+c4v4+c5v5)e3t with v1=[80−310]T, v2=[10003]T, v3=[3−2−100]T, v4=[2−20−30]T, v5=[1−1003]T
33. x1(t)=[cos 4tsin 4t00]T e3t, x2(t)=[−sin 4tcos 4t00]T e3t, x3(t)=[t cos 4tt sin 4tcos 4tsin 4t]T e3t, x4(t)=[−t sin 4tt cos 4t−sin 4tcos 4t]T e3t
34. x1(t)=[sin 3t3 cos 3t−3 cos 3t0sin 3t]e2t, x2(t)=[−cos 3t3 cos 3t+3 cos 3t0−cos 3t]e2t, x3(t)=[3 cos 3t+t sin 3t(3t−10) cos 3t−(3t+9) sin 3tsin 3tt sin 3t]e2t, x4(t)=[−t cos 3t+t sin 3t(3t+9) cos 3t+(3t−10) sin 3t−cos 3t−t cos 3t]e2t
35. x1(t)=x2(t)=v0(1−e−t); limt→∞x1(t)=limt→∞x2(t)=v0
36. x1(t)=v0(2−2e−t−te−t), x2(t)=v0(2−2e−t−te−t−12t2e−t); limt→∞ x1(t)=limt→∞ x2(t)=2v0

In Problems 37, 39, 41, 43, and 45 we give a nonsingular matrix Q and a Jordan-form matrix J such that A=QJQ−1. Any scalar multiple of Q will do the same job.

37. Q=[14−2−109290], J=[−10−0030003]
39. Q=[14−1010100], J=[210021002]
41. Q=[0−50130181], J=[200021002]
43. Q=[−10003−1−1210102001], J=[−11000−10000210002]
45. Q=[−304323600−77222809−216−108030−432−28812], J=[1000011000110001]

Section 7.7

The format for the first eight answers is this: (x(t), y(t)) at t=0.2 by the Euler method, by the improved Euler method, by the Runge-Kutta method, and finally the actual values.

1. (0.8800, 2.5000), (0.9600, 2.6000), (1.0027, 2.6401), (1.0034, 2.6408)
2. (0.8100,−0.8100), (0.8200,−0.8200), (0.8187,−0.8187), (0.8187,−0.8187)
3. (2.8100, 2.3100), (3.2200, 2.6200), (3.6481, 2.9407), (3.6775, 2.9628)
4. (3.3100,−1.6200), (3.8200,−2.0400), (4.2274,−2.4060), (4.2427,−2.4205)
5. (−0.5200, 2.9200), (−0.8400, 2.4400), (−0.5712, 2.4485), (−0.5793, 2.4488)
6. (−1.7600, 4.6800), (−1.9200, 4.5600), (−1.9029, 4.4995), (−1.9025, 4.4999)
7. (3.1200, 1.6800), (3.2400, 1.7600), (3.2816, 1.7899), (3.2820, 1.7902)
8. (2.1600,−0.6300), (2.5200,−0.4600), (2.5320,−0.3867), (2.5270,−0.3889)
9. At t=1 we obtain (x, y)=(3.99261, 6.21770)(h=0.1) and (3.99234, 6.21768) (h=0.05); the actual value is (3.99232, 6.21768).
10. At t=1 we obtain (x, y)=(1.31498, 1.02537)(h=0.1) and (1.31501, 1.02538) (h=0.05);the actual value is (1.31501, 1.02538).
11. At t=1 we obtain (x, y)=(−0.05832, 0.56664)(h=0.1) and (−0.05832, 0.56665) (h=0.05);the actual value is (−0.05832, 0.56665).
12. We solved x′=y, y′=−x+ sin t, x(0)=y(0)=0. With h=0.1 and also with h=0.05 we obtain theactual value x(1.0)≈0.15058.
13. Runge-Kutta, h=0.1: about 1050 ft in about 7.7 s
14. Runge-Kutta, h=0.1: about 1044 ft in about 7.8 s
15. Runge-Kutta, h=1.0: about 83.83 mi in about 168 s
16. At 40°: 5.0 s, 352.9 ft; at 45°: 5.4 s, 347.2 ft; at 50°: 5.8 s, 334.2 ft (all values approximate)
17. At 39.0° the range is about 352.7 ft. At 39.5° it is 352.8; at 40°, 352.9; at 40.5°, 352.6; at 41.0°, 352.1.
18. Just under 57.5°
19. Approximately 253 ft/s
20. Maximum height: about 1005 ft, attained in about 5.6 s; range: about 1880 ft; time aloft: about 11.6 s
21. Runge-Kutta with h=0.1 yields these results: (a) 21400 ft, 46 s, 518 ft/s; (b) 8970 ft, 17.5 s;(c) 368 ft/s (at t≈23).

Section 8.1

1. Φ(t)=[ete3t−ete3t], x(t)=12[5et+e3t−5et+e3t]
2. Φ(t)=[1e4t2−2e4t], x(t)=14[3+5e4t6−10e4t]
3. Φ(t)=[5 cos 4t−5 sin 4t2 cos 4t+4 sin 4t4 cos 4t−2 sin 4t], x(t)=14[−5 sin 4t4 cos 4t−2 sin 4t]
4. Φ(t)=e2t[11+t1t], x(t)=e2t[1+tt]
5. Φ(t)=[2 cos 3t−2 sin 3t−3 cos 3t+3 sin 3t3 cos 3t+3 sin 3t], x(t)=13[3 cos 3t−sin 3t−3 cos 3t+6 sin 3t]
6. Φ(t)=e5t[cos 4t−2 sin 4t2 cos 4t+2 sin 4t2 cos 4t2 sin 4t], x(t)=2e5t[cos 4t+sin 4tsin 4t]
7. Φ(t)=[63et2e−t2ete−t52et2e−t], x(t)=[−12+12et+2e−t−4+4et+e−t−10+8et+2e−t]
8. Φ(t)=[0ete3te−2t−et−e3t−e−2t0e3t], x(t)=[et−et+e−2t−e−2t]
9. eAt=[2e3t−et−2e3t+2ete3t−et−e3t+2et]
10. eAt=[−2+3e2t3−3e2t−2+2e2t3−2e2t]
11. eAt=[3e3t−2e2t−3e3t+3e2t2e3t−2e2t−2e3t+3e2t]
12. eAt=[−3et+4e2t4et−4e2t−3et+3e2t4et−3e2t]
13. eAt=[4e3t−3et−4e3t+4et3e3t−3et−3e3t+4et]
14. eAt=[−8et+9e2t6et−6e2t−12et+12e2t9et−8e2t]
15. eAt=[5e2t−4et−10e2t+10et2e2t−2et−4e2t+5et]
16. eAt=[−9et+10e2t15et−15e2t−6et+6e2t10et−9e2t]
17. eAt=12[e4t+e2te4t−e2te4t−e2te4t+e2t]
18. eAt=12[e2t+e6t−e2t+e6t−e2t+e6te2t+e6t]
19. eAt=15[4e10t+e5t2e10t−2e5t2e10t−2e5te10t+4e5t]
20. eAt=15[e5t+4e15t−2e5t+2e15t−2e5t+2e15t4e5t+e15t]
21. eAt=[1+t−tt1−t]
22. eAt=[1+6t4t−9t1−6t]
23. eAt=[1+t−t−t−t2t1−tt−t2001]
24. eAt=[1+3t0−3t5t+18t217t−18t23t01−3t]
25. eAt=[e2t5te2t0e2t], x(t)=eAt[47]
26. eAt=[e7t011te7te7t], x(t)=eAt[5−10]
27. eAt=[et2tet(3t+2t2)et0et2tet00et], x(t)=eAt[456]
28. eAt=[e5t0010te5te5t0(20t+150t)e5t30te5te5t], x(t)=eAt[405060]
29. eAt=[12t3t+6t24t+6t2+4t3016t3t+6t20012t0001] et, x(t)=eAt[1111]
30. eAt=e3t[10006t1009t+18t26t1012t+54t2+36t39t+18t26t1], x(t)=eAt[1111]
33. x(t)=[c1 cosh t+c2 sinh tc1 sinh t+c2 cosh t]
35. eAt=[e3t4te3t0e3t]
36. eAt=et[12t3t+4t2014t001]
37. eAt=[e2t3e2t−3et13e2t−(13+9t)et0et3tet00et]
38. eAt=[e5t4e10t−4e5t16e10t−(16+50t)e5t0e10t4e10t−4e5t00e5t]
39. eAt=[et3tet12e2t−(12+9t)et(51+18t)et−(51−36t)e2t0et3e2t−3et6et−(6−9t)e2t00e2t4e3t−4e2t000e2t]
40. eAt=[e2t4te2t(4t+8t2)e2t100e3t−(100+96t+32t2)e2t0e2t4te2t20e3t−(20+16t)e2t00e2t4e3t−4e2t000e3t]

Section 8.2

1. x(t)=73, y(t)=−83
2. x(t)=18(1+12t), y(t)=−14(5+4t)
3. x(t)=1756(864e−t+4e6t−868+840t−504t2), y(t)=1756(−864e−t+3e6t+861−882t+378t2)
4. x(t)=184(99e5t−8e−2t−7et), y(t)=184(99e5t+48e−2t−63et)
5. x(t)=13(−12−e−t−7te−t), y(t)=13(−6−7te−t)
6. x(t)=−1256(91+16t)et, y(t)=132(25+16t)et
7. x(t)=1410(369et+166e−9t−125 cos t−105 sin t), y(t)=1410(369et−249e−9t−120 cos t−150 sin t)
8. x(t)=13(17 cos t+2 sin t), y(t)=13(3 cos t+5 sin t)
9. x(t)=14(sin 2t+2t cos 2t+t sin 2t), y(t)=14t sin 2t
10. x(t)=113et(4 cos t−6 sin t), y(t)=113et(3 cos t+2 sin t)
11. x(t)=12(1−4t+e4t), y(t)=14(−5+4t+e4t)
12. x(t)=t2, y(t)=−t2
13. x(t)=12(1+5t)et, y(t)=−52tet
14. x(t)=18(−2+4t−e4t+2te4t), y(t)=12t(−2+e4t)
15. (a) x1(t)=200(1−e−t/10), x2(t)=400(1+e−t/10−2e−t/20) (b) x1(t)→200 and x2(t)→400 as t→+∞ (c) Tank 1: about 6 min 56 s; tank 2: about 24 min 34 s
16. (a) x1(t)=600(1−e−t/20), x2(t)=300(1+e−t/10−2e−t/20) (b) x1(t)→600 and x2(t)→300 as t→∞ (c) Tank 1: about 8 min 7 sec; tank 2: about 17 min 13 sec
17. x1(t)=102−95e−t−7e5t, x2(t)=96−95e−t−e5t
18. x1(t)=68−110t−75e−t+7e5t, x2(t)=74−80t−75e−t+e5t
19. x1(t)=−70−60t+16e−3t+54e2t, x2(t)=5−60t−32e−3t+27e2t
20. x1(t)=3e2t+60te2t−3e−3t, x2(t)=−6e2t+30te2t+6e−3t
21. x1(t)=−e−t−14e2t+15e3t, x2(t)=−5e−t−10e2t+15e3t
22. x1(t)=−10e−t−7te−t+10e3t−5te3t, x2(t)=−15e−t−35te−t+15e3t−5te3t
23. x1(t)=3+11t+8t2, x2(t)=5+17t+24t2
24. x1(t)=2+t+ ln t, x2(t)=5+3t−1t+3 ln t
25. x1(t)=−1+8t+ cos t−8 sin t, x2(t)=−2+4t+2 cos t−3 sin t
26. x1(t)=3 cos t−32 sin t+17t cos t+4t sin t, x2(t)=5 cos t−13 sin t+6t cos t+5t sin t
27. x1(t)=8t3+6t4, x2(t)=3t2−2t3+3t4
28. x1(t)=−7+14t−6t2+4t2 ln t, x2(t)=−7+9t−3t2+ ln t−2t ln t+2t2 ln t
29. x1(t)=t cos t−(ln cos t)(sin t), x2(t)=t sin t+(ln cos t)(cos t)
30. x1(t)=12t2 cos 2t, x2(t)=12t2 sin 2t
31. x1(t)=(9t2+4t3)et, x2(t)=6t2et, x3(t)=6tet
32. x1(t)=(44+18t)et+(−44+26t)e2t, x2(t)=6et+(−6+6t)e2t, x3(t)=2te2t
33. x1(t)=15t2+60t3+95t4+12t5, x2(t)=15t2+55t3+15t4, x3(t)=15t2+20t3, x4(t)=15t2
34. x1(t)=4t3+(4+16t+8t2)e2t, x2(t)=3t2+(2+4t)e2t, x3(t)=(2+4t+2t2)e2t, x4(t)=(1+t)e2t

Section 8.3

1. x(t)=12[e−t+e3t−e−t+e3t−e−t+e3te−t+e3t]
2. x=15[2e−t+3e4t−3e−t+3e4t−2e−t+2e4t3e−t+2e4t]
3. x(t)=17[3e−t+4e6t−4e−t+4e6t−3e−t+3e6t4e−t+3e6t]
4. x(t)=17[e−2t+6e5t−e−2t+e5t−6e−2t+6e5t6e−2t+e5t]
5. x(t)=16[−e−t+7e5t7e−t−7e5t−e−t+e5t7e−t−e5t]
6. x(t)=[−5e3t+6e4t−5e3t+5e4t6e3t−6e4t6e3t−5e4t]
7. x(t)=15[2e−9t+3et−2e−9t+2et−3e−9t+3et3e−9t+2et]
8. x(t)=12[2 cos 2t+sin 2t−5 sin 2tsin 2t2 cos 2t−sin 2t]
9. x(t)=14[4 cos 4t+2 sin 4t−5 sin 4t4 sin 4t4 cos 4t−2 sin 4t]
10. x(t)=13[3 cos 3t−3 sin 3t−2 sin 3t9 sin 3t3 cos 3t+3 sin 3t]
11. x(t)=et[cos 2t−sin 2tsin 2tcos 2t]
12. x(t)=e2t2[2 cos 2t−sin 2t−5 sin 2tsin 2t2 cos 2t+sin 2t]
13. x(t)=e2t3[3 cos 3t+3 sin 3t−9 sin 3t2 sin 3t3 cos 3t−3 sin 3t]
14. x(t)=e3t[cos 4t−sin 4tsin 4tcos 4t]
15. x(t)=e5t4[4 cos 4t+2 sin 4t−5 sin 4t4 sin 4t4 cos 4t−2 sin 2t]
16. x(t)=19[4e−100t+5e−10t−2e−100t+2e−10t−10e−100t+10e−10t5e−100t+4e−10t]
17. x(t)=16[3+e6t+2e9t−2e6t+2e9t−3+e6t+2e9t−2e6t+2e9t4e6t+2e9t−2e6t+2e9t−3+e6t+2e9t−2e6t+2e9t3+e6t+2e9t]
18. x(t)=118[16+e6t+2e9t−4+4e9t−4+4e9t−4e6t+4e9t1+9e6t+8e9t1−9e6t+8e9t−4e6t+4e9t1−9e6t+8e9t1+9e6t+8e9t]
19. x(t)=13[2e3t+e6t−e3t+e6t−e3t+e6t−e3t+e6t2e3t+e6t−e3t+e6t−e3t+e6t−e3t+e6t2e3t+e6t]
20. x(t)=16[3e2t+e6t+2e9t−2e6t+2e9t−3e2t+e6t+2e9t−2e6t+e9t4e6t+2e9t−2e6t+e9t−3e2t+e6t+2e9t−2e6t+e9t3e2t+e6t+2e9t]

21. x(t)=e−3t[1+tt−t1−t]
22. x(t)=e2t[1+t−tt1−t]
23. x(t)=e3t[1−2t−2t2t1+2t]
24. x(t)=e4t[1−t−tt1+t]
25. x(t)=e5t[1+2tt−4t1−2t]
26. x(t)=e5t[1−4t−4t4t1+4t]
27. x(t)=[e2t00e2t−e9te9t−e2t+e9t00e2t]
28. x(t)=[−2e7t+3e13t−2e7t+2e13t03e7t−3e13t3e7t−2e13t0−e7t+e13t−e7t+e13te13t]
29. x(t)=[7e5t−6e9t−3e5t+3e9t−21e5t+21e9t0e5t02e5t−2e9t−e5t+e9t−6e5t+7e9t]
30. x(t)=[5e3t−4e7t−10e3t+10e7t12e3t−12e7t2e3t−2e7t−4e3t+5e7t6e3t−6e7t00e3t]
31. eAt=14[−e−t+5e3te−t−e3t−5e−t+5e3t5e−t−e3t], x(t)=[−e−t−14e2t+15e3t−5e−t−10e2t+15e3t]
32. With eAt as in Problem 31 , x(t)=[(−10−7t)e−t+(10−5t)e3t(−15−35t)e−t+(15−5t)e3t].
33. eAt=[1+3t−t9t1−3t], x(t)=[3+11t+8t25+17t+24t2]
34. With eAt as in Problem 33 , x(t)=[2+t+ln t5+3t−1t+3 ln t].
35. eAt=[cos t+ 2 sin t−5 sin tsin tcos t− 2 sin t], x(t)=[−1+8t+cos t−8 sin t−2+4t+2 cos t−3 sin t]
36. With eAt as in Problem 35 , x(t)=[3 cos t−32 sin t+17t cos t+4t sin t5 cos t−13 sin t+6t cos t+5t sin t].
37. eAt=[1+2t−4tt1−2t], x(t)=[8t3+6t43t2−2t3+3t4]
38. With eAt as in Problem 37 , x(t)=[−7+14t−6t2+4t2 ln t−7+9t−3t2+ln t−2t ln t+2t2 ln t].
39. eAt=[cos t−sin tsin tcos t], x(t)=[t cos t−(ln cos t)(sin t)t sin t+(ln cos t)(cos t)]
40. eAt=[cos 2t−2 sin 2tsin 2tcos 2t], x(t)=[12t2 cos 2t12t2 sin 2t]
41. x(t)=[−9e−t+10e3t−2e−t+2e3t4e−t−4e3t9e−t−9e3t2e−t−2e3t−4e−t+4e3t−18e−t+18e3t−4e−t+4e3t8e−t−7e3t]
42. x(t)=[−5e−2t+6e3t−10e−2t+10e3t−20e−2t+20e3t−3e−2t+3e3t−6e−2t+7e3t−12e−2t+12e3t3e−2t+−3e3t6e−2t−6e3t12e−2t−11e3t]
43. x(t)=12e2t[−t2−8t+24t2+34tt2+8t−2t8t+22t−t24t2+2tt2+2]
44. x(t)=12e3t[4t+2−2t2t2t2+2t−t2+2t22t2−6t−t2+4tt2−4t+2]
45. x(t)=[e−tte−tte−t−2te−t−3e−t+(3−2t)e2t(1−3t)e−t(1−3t)e−t−e2t2(3t−1)e−t+(2−t)e2t−e−t+(1+2t)e2t−te−t−te−t+e2t2te−t+te2t−2e−t+2e2t−2te−t−2te−t4te−t+e2t]
46. x(t)=12et[48t2+68t+2−18t2−24t6t2+8t36t2+60t7t2+44t−3t2−18t+2t2+6t6t2+38t−21t2−20t9t2+6t−3t2−2t+2−18t2−18t−42t2−54t18t2+18t−6t2−6t−36t2−48t+2]
47. x(t)=[cos tcos t]+i[−sin 3tsin 3t] There are two natural modes—one in which the two masses move in the same direction with frequency ω1=1 and with equal amplitudes, and one in which they move in opposite directions with frequency ω2=3 and with equal amplitudes.
48. x(t)=[cos tcos t]+i[−sin 5 tsin 5 t] There are two natural modes—one in which the two masses move in the same direction with frequency ω1=1 and with equal amplitudes, and one in which they move in opposite directions with frequency ω2=5 and with equal amplitudes.
49. x(t)=[cos 2 tcos 2 t]+i[−sin 2tsin 2t] There are two natural modes—one in which the two masses move in the same direction with frequency ω1=2 and with equal amplitudes, and one in which they move in opposite directions with frequency ω2=2 and with equal amplitudes.
50. x(t)=[cos 2tcos 2t]+i[−sin 4tsin 4t] There are two natural modes-one in which the two masses move in the same direction with frequency ω1=2 and with equal amplitudes, and one in which they move in opposite directions with frequency ω2=4 and with equal amplitudes.

Section 9.1

1. 6.1.14
2. 6.1.16
3. 6.1.19
4. 6.1.13
5. 6.1.12
6. 6.1.18
7. 6.1.15
8. 6.1.17
9. Equilibrium solutions x(t)≡0, ±2. The critical point (0, 0) in the phase plane looks like a center, whereas the points (±2, 0) look like saddle points.
10. Equilibrium solution x(t)≡0. The critical point (0, 0) in the phase plane looks like a spiral sink.
11. Equilibrium solutions x(t)≡…,−2π,−π, 0, π, 2π,…. The phase portrait shown in the solutions manual suggests that the critical point (nπ, 0) in the phase plane is a spiral sink if n is even, but is a saddle point if n is odd.
12. Equilibrium solution x(t)≡0. The critical point (0, 0) in the phase plane looks like a spiral source, with the solution curves emanating from this source spiraling outward toward a closed curve trajectory.
13. Solution x(t)=x0e−2t, y(t)=y0e−2t. The origin is a stable proper node similar to the one illustrated in Fig. 6.1.4.
14. Solution x(t)=x0e2t, y(t)=y0e−2t. The origin is an unstable saddle point.
15. Solution x(t)=x0e−2t, y(t)=y0e−t. The origin is a stable node.
16. Solution x(t)=x0et, y(t)=y0e3t. The origin is an unstable improper node.
17. Solution x(t)=A cos t+B sin t, y(t)=B cos t−A sin t. The origin is a stable center.
18. Solution x(t)=A cos 2t+B sin 2t, y(t)=−2B cos 2t+2A sin 2t. The origin is a stable center.
19. Solution x(t)=A cos 2t+B sin 2t, y(t)=B cos 2t−A sin 2t. The origin is a stable center.
20. Solution x(t)=e−2t(A cos t+B sin t), y(t)=e−2t[(−2A+B) cos t−(A+2B) sin t]. The origin is a stable spiral point.
23. The origin and the circles x2+y2=C>0; the origin is a stable center.
24. The origin and the hyperbolas y2−x2=C; the origin is an unstable saddle point.
25. The origin and the ellipses x2+4y2=C>0; the origin is a stable center.
26. The origin and the ovals of the form x4+y4=C>0; the origin is a stable center.

Section 9.2

1. Asymptotically stable node
2. Unstable improper node
3. Unstable saddle point
4. Unstable saddle point
5. Asymptotically stable node
6. Unstable node
7. Unstable spiral point
8. Asymptotically stable spiral point
9. Stable, but not asymptotically stable, center
10. Stable, but not asymptotically stable, center
11. Asymptotically stable node: (2, 1)
12. Unstable improper node: (2,−3)
13. Unstable saddle point: (2, 2)
14. Unstable saddle point: (3, 4)
15. Asymptotically stable spiral point: (1, 1)
16. Unstable spiral point: (3, 2)
17. Stable center: (52,−12)
18. Stable, but not asymptotically stable, center: (−2,−1)
19. (0, 0) is a stable node. Also, there is a saddle point at (0.67, 0.40).
20. (0, 0) is an unstable node. Also, there is a saddle point at (−1,−1) and a spiral sink at (−2.30,−1.70).
21. (0, 0) is an unstable saddle point. Also, there is a spiral sink at (−0.51,−2.12).
22. (0, 0) is an unstable saddle point. Also, there are nodal sinks at (±0.82,±5.06) and nodal sources at (±3.65,∓0.59).
23. (0, 0) is a spiral sink. Also, there is a saddle point at (−1.08,−0.68).
24. (0, 0) is a spiral source. No other critical points are visible.
25. Theorem 2 implies only that (0, 0) is a stable sink—either a node or a spiral point. The phase portrait for −5≤x, y≤5 also shows a saddle point at (0.74,−3.28) and spiral sink at (2.47,−0.46). The origin looks like a nodal sink in a second phase portrait for −0.2≤x, y≤0.2, which also reveals a second saddle point at (0.12, 0.07).
26. Theorem 2 implies only that (0, 0) is an unstable source. The phase portrait for −3≤x, y≤3 also shows saddle points at (0.20, 0.25) and (−0.23,−1.50), as well as a nodal sink at (2.36, 0.58).
27. Theorem 2 implies only that (0, 0) is a center or a spiral point, but does not establish its stability. The phase portrait for −2≤x, y≤2 also shows saddle points at (−0.25,−0.51) and (−1.56, 1.64), plus a nodal sink at (−1.07,−1.20). The origin looks like a likely center in a second phase portrait for −0.6≤x, y≤0.6.
28. Theorem 2 implies only that (0, 0) is a center or a spiral point, but does not establish its stability (though in the phase portrait it looks like a likely center). The phase portrait for −0.25≤x≤0.25, −1≤y≤1 also shows saddle points at (0.13, 0.63) and (−0.12,−0.47).
29. There is a saddle point at (0, 0). The other critical point (1, 1) is indeterminate, but looks like a center in the phase portrait.
30. There is a saddle point at (1, 1) and a spiral sink at (−1, 1).
31. There is a saddle point at (1, 1) and a spiral sink at (−1,−1).
32. There is a saddle point at (2, 1) and a spiral sink at (−2,−1).
37. Note that the differential equation is homogeneous.

Section 9.3

1. Linearization at (0, 0): x′=200x, y′=−150y; phase plane portrait:
3. Linearization at (75, 50): u′=−300v, v′=100u; phase plane portrait:
5. The characteristic equation is λ2+45λ+126=0.
7. The characteristic equation is (−24−λ)2−2·(18)2=0. Phase plane portrait:

Phase plane portrait for the nonlinear system in Problems 4-7:
9. The characteristic equation is λ2+58λ−120=0.
10. The characteristic equation is (λ+36)(λ+18)−576=0. Phase plane portrait:

Phase plane portrait for the nonlinear system in Problems 8-10:
12. The characteristic equation is λ2+2λ−15=0.
13. The characteristic equation is λ2+2λ+6=0. Phase plane portrait:
15. The characteristic equation is λ2+2λ−24=0.
17. The characteristic equation is λ2−4λ+6=0. Phase plane portrait:
19. The characteristic equation is λ2+10=0. Phase plane portrait:
21. The characteristic equation is λ2−λ−6=0.
22. The characteristic equation is λ2−5λ+10=0. Phase plane portrait:
24. The characteristic equation is λ2+5λ−14=0.
25. The characteristic equation is λ2+5λ+10=0. Phase plane portrait:
26. Naturally growing populations in competition Critical points: nodal source (0, 0) and saddle point (3, 2) Nonzero coexisting populations x(t)≡3, y(t)≡2
27. Naturally declining populations in cooperation Critical points: nodal sink (0, 0) and saddle point (3, 2) Nonzero coexisting populations x(t)≡3, y(t)≡2
28. Naturally declining predator, naturally growing prey population Critical points: saddle point (0, 0) and apparent stable center (4, 8)

Nonzero coexisting populations x(t)≡4, y(t)≡8
29. Logistic and naturally growing populations in competition Critical points: nodal source (0, 0), nodal sink (3, 0), and saddle point (2, 2) Nonzero coexisting populations x(t)≡2, y(t)≡2
30. Logistic and naturally declining populations in cooperation Critical points: saddle point (0, 0), nodal sink (3, 0) and saddle point (5, 4) Nonzero coexisting populations x(t)≡5, y(t)≡4
31. Logistic prey, naturally declining predator population Critical points: saddle points (0, 0) and (3, 0), spiral sink (2, 4) Nonzero coexisting populations x(t)≡2, y(t)≡4
32. Logistic populations in cooperation

Critical points: nodal source (0, 0), saddle points (10, 0) and (0, 20), nodal sink (30, 60)

Nonzero coexisting populations x(t)≡30, y(t)≡60
33. Logistic prey and predator populations Critical points: nodal source (0, 0), saddle points (30, 0) and (0, 20), nodal sink (4, 22)

Nonzero coexisting populations x(t)≡4, y(t)≡22
34. Logistic prey and predator populations Critical points: nodal source (0, 0), saddle points (15, 0) and (0, 5), spiral sink (10, 10)

Nonzero coexisting populations x(t)≡10, y(t)≡10

Section 9.4

1. Eigenvalues: −2, −3; stable node
2. Eigenvalues: 1, 3; unstable node
3. Eigenvalues: −3, 5; unstable saddle point
4. Eigenvalues: −1±2i; stable spiral point
5. Critical points: (0, nπ) where n is an integer; an unstable saddle point if n is even, a stable spiral point if n is odd
6. Critical points: (n, 0) where n is an integer; an unstable saddle point if n is even, a stable spiral point if n is odd
7. Critical points: (nπ, nπ) where n is an integer; an unstable saddle point if n is even, a stable spiral point if n is odd
8. Critical points: (nπ, 0) where n is an integer; an unstable nodeif n is even, an unstable saddle point if n is odd
9. If n is odd then (nπ, 0) is an unstable saddle point.
10. If n is odd then (nπ, 0) is a stable node.
11. (nπ, 0) is a stable spiral point.
12. Unstable saddle points at (2, 0) and (−2, 0), a stable center at (0, 0)
13. Unstable saddle points at (2, 0) and (−2, 0), a stable spiral point at (0, 0)
14. Stable centers at (2, 0) and (−2, 0), an unstable saddle point at (0, 0)
15. A stable center at (0, 0) and an unstable saddle point at (4, 0)
16. Stable centers at (2, 0), (0, 0) and (−2, 0), unstable saddle points at (1, 0) and (−1, 0)
17. (0, 0) is a spiral sink.
18. (0, 0) is a spiral sink; the points (±2, 0) are saddle points.
19. (0, 0) is a spiral sink.
20. (nπ, 0) is a spiral sink if n is even, a saddle point if n is odd.

Section 10.1

1. 1/s2, s>0
2. 2/s3, s>0
3. e/(s−3), s>3
4. s/(s2+1), s>0
5. 1/(s2−1), s>1
6. 12[1/s−s/(s2+4)], s>0
7. (1−e−s)/s, s>0
8. (e−s−e−2s)/s, s>0
9. (1−e−s−se−s)/s2, s>0
10. (s−1+e−s)/s2, s>0
11. 12πs−3/2+3s−2, s>0
12. (45π−192s3/2)/(8s7/2), s>0
13. s−2−2(s−3)−1, s>3
14. 3π/(4s5/2)+1/(s+10), s>0
15. s−1+s(s2−25)−1, s>5
16. (s+2)/(s2+4), s>0
17. cos2 2t=12(1+ cos 4t); 12[s−1+s/(s2+16)], s>0
18. 3/(s2+36), s>0
19. s−1+3s−2+6s−3+6s−4, s>0
20. 1/(s−1)2, s>1
21. (s2−4)/(s2+4)2, s>0
22. 12[s/(s2−36)−s−1]
23. 12t3
24. 2t/π
25. 1−83t3/2π−1/2
26. e−5t
27. 3e4t
28. 3 cos 2t+12sin 2t
29. 53sin 3t−3 cos 3t
30. −cosh 2t−92 sinh 2t
31. 35 sinh 5t−10 cosh 5t
32. 2u(t−3)
37. f(t)=1−u(t−a). Your figure should indicate that the graph of f contains the point (a, 0), but not the point (a, 1).
38. f(t)=u(t−a)−u(t−b). Your figure should indicate that the graph of f contains the points (a, 1) and (b, 0), but not the points (a, 0) and (b, 1).
39. Figure 10.2.8 shows the graph of the unit staircase function.

Section 10.2

1. x(t)=5 cos 2t
2. x(t)=3 cos 3t+43sin 3t
3. x(t)=23(e2t−e−t)
4. x(t)=12(7e−3t−3e−5t)
5. x(t)=13(2 sin t−sin 2t)
6. x(t)=13(cos t− cos 2t)
7. x(t)=18(9 cos t−cos 3t)
8. x(t)=19(1− cos 3t)
9. x(t)=16(2−3e−t+e−3t)
10. x(t)=14(2t−3+12e−t−9e−2t)
11. x(t)=1, y(t)=−2
12. x(t)=29(e2t−e−t−3te−t), y(t)=19(e2t−e−t+6te−t)
13. x(t)=−(2/3) sinh (t/3), y(t)=cosh (t/3)+(1/3) sinh (t/3)
14. x(t)=14(2t−3 sin 2t), y(t)=−18(2t+3 sin 2t)
15. x(t)=13(2+e−3t/2 [cos (rt/2)+r sin (rt/2)]), y(t)=121(28−9et+2e−3t/2 [cos (rt/2)+4r sin (rt/2)]) where r=3
16. x(t)=cos t+ sin t, y(t)=et− cos t, z(t)=−2 sin t
17. f(t)=13(e3t−1)
18. f(t)=35(1−e−5t)
19. f(t)=14(1− cos 2t)=12sin2 t
20. f(t)=19(6 sin 3t− cos 3t+1)
21. f(t)=t− sin t
22. f(t)=19(−1+ cosh 3t)
23. f(t)=−t+ sinh t
24. f(t)=12(e−2t−2e−t+1)

Section 10.3

1. 24/(s−π)5
2. 34π (s+4)−5/2
3. 3π/[(s+2)2+9π2]
4. 2 (2s+5)/(4s2+4s+17)
5. 32e2t
6. (t−t2)e−t
7. te−2t
8. e−2t cos t
9. e3t(3 cos 4t +72sin 4t)
10. 136e2t/3(8 cos 43t−5 sin 43t)
11. 12 sinh 2t
12. 2+3e3t
13. 3e−2t−5e−5t
14. 2+e2t−3e−t
15. 125(e5t−1−5t)
16. 1125[e2t(5t−2)+e−3t(5t+2)]
17. 116(sinh 2t−sin 2t)
18. e4t(1+12t+24t2+323t3)
19. 13(2 cos 2t+2 sin 2t−2 cos t− sin t)
20. 132[e2t(2t−1)+e−2t(2t+1) ]
21. 12e−t(5 sin t−3t cos t−2t sin t)
22. 164et/2[(4t+8) cos t+(4−3t) sin t]
27. 14e−3t(8 cos 4t+9 sin 4t)
28. 14(1−2e2t+e4t)
29. 18(−6t+3 sinh 2t)
30. 110[2e−t−e−2t(2 cos 2t+ sin 2t)]
31. 115(6e2t−5−e−3t)
32. 12(cosh t+ cos t)
33. x(t)=r(cosh rt sin rt− sinh rt cos rt) where r=1/2
34. 12sin 2t+13sin 3t
35. 116(sin 2t−2t cos 2t)
36. 150[2e2t+(10t−2) cos t−(5t+14) sin t]
37. 150[(5t−1)e−t+e−2t(cos 3t+32 sin 3t)]
38. 1510e−3t(489 cos 3t+307 sin 3t)+ 1170(7 cos 2t+6 sin 2t)
39.

Section 10.4

1. 12t2
2. (eat−at−1)/a2
3. 12(sin t−t cos t)
4. 2(t− sin t)
5. teat
6. (eat−ebt)/(a−b)
7. 13(e3t−1)
8. 14(1− cos 2t)
9. 154(sin 3t−3t cos 3t)
10. (kt− sin kt)/k3
11. 14(sin 2t+2t cos 2t)
12. 15[1−e−2t(cos t+2 sin t)]
13. 110(3e3t−3 cos t+sin t)
14. 13(cos t− cos 2t)
15. 6s/(s2+9)2, s>0
16. (2s3−24s)/(s2+4)3, s>0
17. (s2−4s−5)/(s2−4s+13)2, s>0
18. 2(3s2+6s+7)(s+1)2(s2+2s+5)2, s>0
19. 12π−arctans=arctan(1/s), s>0
20. 12ln(s2+4)−ln s, s>0
21. ln s−ln(s−3), s>3
22. ln (s+1)− ln (s−1), s>1
23. −(2 sinh 2t)/t
24. 2(cos 2t− cos t)/t
25. e−2t+e3t−2 cos t)/t
26. (e−2t sin 3t)/t
27. 2(1−cos t)/t
28. 18(t sin t−t2 cos t)
29. (s+1)X′(s)+4X(s)=0; x(t)=Ct3e−t, C≠0
30. X(s)=A/(s+3)3; x(t)=Ct2e−3t, C≠0
31. (s−2)X′(s)+3X(s)=0; x(t)=Ct2e2t, C≠0
32. (s2+2s)X′(s)+(4s+4)X(s)=0; x(t)=C(1−t−e−2t−te−2t), C≠0
33. (s2+1)X′(s)+4sX(s)=0; x(t)=C(sin t−t cos t), C≠0
34. x(t)=Ce−2t(sin 3t−3t cos 3t), C≠0

Section 10.5

1. f(t)=u(t−3)·(t−3)
2. f(t)=(t−1)u(t−1)−(t−3)u(t−3)
3. f(t)=u(t−1)·e−2(t−1)
4. f(t)=et−1u(t−1)−e2et−2u(t−2)
5. f(t)=u(t−π)· sin (t−π)=−u(t−π) sin t
6. f(t)=u(t−1)· cos π(t−1)=−u(t−1) cos πt
7. f(t)=sin t−u(t−2π) sin (t−2π)=[1−u(t−2π)] sin t
8. f(t)=cos πt−u(t−2) cos π(t−2)=[1−u(t−2)] cos πt
9. f(t)=cos πt+u(t−3) cos π(t−3)=[1−u(t−3)] cos πt
10. f(t)=2u(t−π) cos 2(t−π)−2u(t−2π) cos 2(t−2π)=2[u(t−π)−u(t−2π)] cos 2t
11. f(t)=2[1−u3(t)]; F(s)=2(1−e−3s)/s
12. F(s)=(e−s−e−4s)/s
13. F(s)=(1−e−2πs)/(s2+1)
14. F(s)=s(1−e−2s)/(s2+π2)
15. F(s)=(1+e−3πs)/(s2+1)
16. F(s)=2(e−πs−e−2πs)/(s2+4)
17. F(s)=π(e−2s+e−3s)/(s2+π2)
18. F(s)=2π(e−3s+e−5s)/(4s2+π2)
19. F(s)=e−s(s−1+s−2)
20. F(s)=(1−e−s)/s2
21. F(s)=(1−2e−s+e−2s)/s2
28. F(s)=(1−e−as−ase−as)/[s2(1−e−2as)]
31. x(t)=12[1−u(t−π)] sin2 t
32. x(t)=g(t)−u(t−2)g(t−2), where g(t)=112(3−4e−t+e−4t)
33. x(t)=18[1−u(t−2π)](sin t−13sin 3t)
34. x(t)=g(t)−u(t−1)[g(t−1)+h(t−1)], where g(t)=t− sin t and h(t)=1− cos t
35. x(t)=14{−1+t+(t+1)e−2t+ u(t−2)[1−t+(3t−5)e−2(t−2)]}
36. x(t)=2| sin t| sin t
37. x(t)=g(t)+2∑n=1∞(−1)nu(t−nπ)g(t−nπ), where g(t)=1−13e−t(3 cos 3t+ sin 3t)

Section 11.1

1. y(x)=c0(1+x+x22+x33!+⋯)=c0ex; ρ=+∞
2. y(x)=c0(1+4x1!+42x22!+43x33!+44x44!+⋯)=c0e4x; ρ=∞
3. y(x)=c0(1−3x2+(3x)22!22−(3x)33!23+(3x)44!24−⋯)=c0e−3x/2; ρ=+∞
4. y(x)=c0(1−x21!+x42!−x63!+⋯)=c0e−x2; ρ=∞
5. y(x)=c0(1+x33+x62!32+x93!33+⋯)=c0exp(13x3); ρ=+∞
6. y(x)=c0(1+x2+x24+x38+x416+⋯)=2c02−x; ρ=2
7. y(x)=c0(1+2x+4x2+8x3+⋯)=c01−2x; ρ=12
8. y(x)=c0(1+x2−x28+x316−5x4128+⋯)=c01+x; ρ=1
9. y(x)=c0(1+2x+3x2+4x3+⋯ )=c0(1−x)2; ρ=1
10. y(x)=c0(1−3x2+3x28+x316+3x4128+⋯) =c0(1−x)3/2; ρ=1
11. y(x)=c0(1+x22!+x44!+x66!+⋯)+ c1(x+x33!+x55!+x77!+⋯) =c0 cosh x+c1 sinh x; ρ=+∞
12. y(x)=c0(1+(2x)22!+(2x)44!+(2x)66!+⋯)+ c12((2x)+(2x)33!+(2x)55!+(2x)77!+⋯) =c0 cosh 2x+c12 sinh 2x; ρ=∞
13. y(x)=c0(1−(3x)22!+(3x)44!−(3x)66!+⋯)+ c13(3x−(3x)33!+(3x)55!−(3x)77!+⋯) =c0 cos 3x+13c1 sin 3x; ρ=+∞
14. y(x)=x+c0(1−x22!+x44!−x66!+⋯) +(c1−1)(x−x33!+x55!−x77!+⋯) =x+c0 cos x+(c1−1) sin x; ρ=∞
15. (n+1)cn=0 for all n≧0, so cn=0 for all n≧0.
16. 2ncn=cn for all n≧0, so cn=0 for all n≧0.
17. c0=c1=0 and cn+1=−ncn for n≧1, thus cn=0 for all n≧0.
18. cn=0 for all n≧0
19. (n+1)(n+2)cn+2=−4cn; y(x)=32[(2x)−(2x)33!+(2x)55!−(2x)77!+⋯]=32sin 2x
20. (n+1)(n+2)cn+2=4cn; y(x)=2[1+(2x)22!+(2x)44!+(2x)66!+⋯]=2 cosh 2x
21. n(n+1)cn+1=2ncn−cn−1; y(x)=x+x2+x32!+x43!+x54!+⋯=xex
22. n(n+1)cn+1=−ncn+2cn−1; y=e−2x
23. As c0=c1=0 and (n2−n+1)cn+(n−1)cn−1=0 for n≧2, cn=0 for all n≧0

Section 11.2

1. cn+2=cn; y(x)=c0∑n=0∞x2n+c1∑n=0∞x2n+1=c0+c1x1−x2; ρ=1
2. cn+2=−12cn; ρ=2; y(x)=c0∑n=0∞(−1)nx2n2n+c1∑n=0∞(−1)nx2n+12n
3. (n+2)cn+2=−cn; y(x)=c0∑n=0∞(−1)nx2nn!2n+c1∑n=0∞(−1)nx2n+1(2n+1)!!; ρ=+∞
4. (n+2)cn+2=−(n+4)cn; ρ=1; c0∑n=0∞(−1)n(n+1)x2n+13c1∑n=0∞(−1)n(2n+3)x2n+1
5. 3(n+2)cn+2=ncn; ρ=3; y(x)=c0+c1∑n=0∞x2n+1(2n+1)3n
6. (n+1)(n+2)cn+2=(n−3)(n−4)cn; ρ=∞; y(x)=c0(1+6x2+x4)+c1(x+x3)
7. 3(n+1)(n+2)cn+2=−(n−4)2cn; y(x)=c0(1−8x23+8x427)+c1(x−x32+x5120+9∑n=3∞(−1)n[(2n−5)!!]2x2n+1(2n+1)!3n)
8. 2(n+1)(n+2)cn+2=(n−4)(n+4)cn; y(x)=c0(1−4x2+2x4)+c1(x−5x34+7x532+∑n=3∞(2n−5)!!(2n+3)!!x2n+1(2n+1)!2n)
9. (n+1)(n+2)cn+2=(n+3)(n+4)cn; ρ=1; y(x)=c0∑n=0∞(n+1)(2n+1)x2n+c13∑n=0∞(n+1)(2n+3)x2n+1
10. 3(n+1)(n+2)cn+2=−(n−4)cn; y(x)=c0(1+2x23+x427)+c1(x+x36+x5360+3∑n=3∞(−1)n(2n−5)!!x2n+1(2n+1)!3n)
11. 5(n+1)(n+2)cn+2=2(n−5)cn; y(x)=c1(x−4x315+4x5375)+c0(1−x2+x410+x6750+15∑n=4∞(2n−7)!!2nx2n(2n)!5n)
12. c2=0; (n+2)cn+3=cn; y(x)=c0(1+∑n=1∞x3n2·5⋯(3n−1))+c1∑n=0∞x3n+1n!3n
13. c2=0; (n+3)cn+3=−cn; y(x)=c0∑n=0∞(−1)nx3nn!3n+c1∑n=0∞(−1)nx3n+11·4⋯(3n+1)
14. c2=0; (n+2)(n+3)cn+3=−cn; y(x)=c0(1+∑n=1∞(−1)nx3n3n·n!·2·5⋯(3n−1))+c1∑n=0∞(−1)nx3n+13n·n!·1·4⋯(3n+1)
15. c2=c3=0; (n+3)(n+4)cn+4=−cn; y(x)=c0(1+∑n=1∞(−1)nx4n4n·n!·3·7⋯(4n−1))+c1∑n=0∞(−1)nx4n+14n·n!·5·9⋯(4n+1)
16. y(x)=x
17. y(x)=1+x2
18. y(x)=2∑n=0∞(−1)n(x−1)2nn!2n; converges for all x
19. y(x)=13∑n=0∞(2n+3)(x−1)2n+1; converges if 0<x<2
20. y(x)=2−6(x−3)2; converges for all x
21. y(x)=1+4(x+2)2; converges for all x
22. y(x)=2x+6
23. 2c2+c0=0; (n+1)(n+2)cn+2+cn+cn−1=0 for n≧1; y1(x)=1−x22−x36+⋯; y2(x)=x−x36−x412+⋯
24. y1(x)=1+x33+x55+x645+⋯; y2(x)=x+x33+x46+x55+⋯
25. c2=c3=0, (n+3)(n+4)cn+4+(n+1)cn+1+cn=0 for n≧0; y1(x)=1−x412+x7126+⋯; y2(x)=x−x412−x520+⋯
26. y(x)=c0(1−x630+x972+⋯)+c1(x−x742+x1090+⋯)
27. y(x)=1−x−x22+x33−x424+x530+29x6720+13x7630−143x840320+⋯; y(0.5)≈0.4156
28. y(x)=c0(1−x22+x36+⋯)+c1(x−x36+x412+⋯)
29. y1(x)=1−12x2+1720x6+⋯; y2(x)=x−16x3−160x5+⋯
30. y(x)=c0(1−x22+x36+⋯)+c1(x−x22+x418+⋯)
33. The following figure shows the interlaced zeros of the 4th and 5th Hermite polynomials.
34. The figure below results when we use n=40 terms in each summation. But with n=50 we get the same picture as Fig. 8.2.3 in the text.

Section 11.3

1. Ordinary point
2. Ordinary point
3. Irregular singular point
4. Irregular singular point
5. Regular singular point; r1=0, r2=−1
6. Regular singular point; r1=1, r2=−2
7. Regular singular point; r=−3, −3
8. Regular singular point; r=12, −3
9. Regular singular point x=1
10. Regular singular point x=1
11. Regular singular points x=1, −1
12. Irregular singular point x=2
13. Regular singular points x=2, −2
14. Irregular singular points x=3, −3
15. Regular singular point x=2
16. Irregular singular point x=0, regular singular point x=1
17. y1(x)=cos x, y2(x)=sin x
18. y1(x)=∑n=0∞xnn!(2n+1)!!, y2(x)=x−1/2∑n=0∞xnn!(2n−1)!!
19. y1(x)=x3/2(1+3∑n=1∞xnn!(2n+3)!!), y2(x)=1−x−∑n=2∞xnn!(2n−3)!!
20. y1(x)=x1/3∑n=0∞(−1)n2nxnn!·4·7⋯(3n+1), y2(x)=∑n=0∞(−1)n2nxnn!⋅2⋅5⋯(3n−1)
21. y1(x)=x(1+∑n=1∞x2nn!·7·11⋯(4n+3)), y2(x)=x−1/2(1+∑n=1∞x2nn!⋅1⋅5⋯(4n−3))
22. y1(x)=x3/2(1+∑n=1∞(−1)nx2nn!·9·13⋯(4n+5)), y2(x)=x−1(1+∑n=1∞(−1)n−1x2nn!⋅3⋅7⋯(4n−1))
23. y1(x)=x1/2(1+∑n=1∞x2n2n·n!·19·31⋯(12n+7)), y2(x)=x−2/3(1+∑n=1∞x2n2n⋅n!⋅5⋅17⋯(12n−7))
24. y1(x)=x1/3(1+∑n=1∞(−1)nx2n2n·n!·7·13⋯(6n+1)), y2(x)=1+∑n=1∞(−1)nx2n2n⋅n!⋅5⋅11⋯(6n+1)
25. y1(x)=x1/2∑n=0∞(−1)nxnn!·2n=x1/2e−x/2, y2(x)=1+∑n=1∞(−1)nxn(2n−1)!!
26. y1(x)=x1/2∑n=0∞x2nn!·2n=x1/2exp(12x2), y2(x)=1+∑n=1∞2nx2n3⋅7⋯(4n−1)
27. y1(x)=1xcos 3x, y2(x)=1xsin 3x
28. y1(x)=1x cosh 2x, y2(x)=1x sinh 2x
29. y1(x)=1xcos x2, y2(x)=1xsin x2
30. y1(x)=cos x2, y2(x)=sin x2
31. y1(x)=x1/2 cosh x, y2(x)=x1/2 sinh x
32. y1(x)=x+x25, y2(x)=x−1/2(1−5x2−15x28−5x348+⋯)
33. y1(x)=x−1(1+10x+5x2+10x39+⋯), y2(x)=x1/2(1+11x20−11x2224+671x324192+⋯)
34. y1(x)=x(1−x242+x41320+⋯), y2(x)=x−1/2(1−7x224+19x43200+⋯)

Section 11.4

5. J4(x)=1x2(x2−24)J0(x)+8x3(6−x2)J1(x)
10. 3
11. x2J1(x)+xJ0(x)−∫J0(x) dx+C
12. (x3−4x)J1(x)+2x2J0(x)+C
13. (x4−9x2)J1(x)+(3x3−9x)J0(x)+9∫J0(x) dx+C
14. −xJ1(x)+∫J0(x) dx+C
15. 2xJ1(x)−x2J0(x)+C
16. 3x2J1(x)+(3x−x3)J0(x)−3∫J0(x) dx+C
17. (4x3−16x)J1(x)+(8x2−x4)J0(x)+C
18. −2J1(x)+∫J0(x) dx+C
19. y(x)=x[c1J0(x)+C2Y0(x)]
20. y(x)=1x[c1J1(x)+c2Y1(x)]
21. y(x)=x[c1J1/2(3x2)+c2J−1/2(3x2)]
22. y(x)=x3[c1J2(2x1/2)+c2Y2(2x1/2)]
23. y(x)=x−1/3[c1J1/3(13x3/2)+c2J−1/3(13x3/2)]
24. y(x)=x−1/4[c1J0(2x3/2)+c2Y0(2x3/2)]
25. y(x)=x−1[c1J0(x)+c2Y0(x)]
26. y(x)=x2[c1J1(4x1/2)+c2Y1(4x1/2)]
27. y(x)=x1/2[c1J1/2(2x3/2)+c2J−1/2(2x3/2)]
28. y(x)=x−1/4[c1J3/2(25x5/2)+c2J−3/2(25x5/2)]
29. y(x)=x1/2[c1J1/6(13x3)+c2J−1/6(13x3)]
30. y(x)=x1/2[c1J1/5(45x5/2)+c2J−1/5(45x5/2)]